Prove that this map is a submersion (check my work please) There are some calculations but I would be very grateful if somebody could check whether my arguments are mathematically sound. I am new to differential geometry. You can also suggest easier solutions.
In the following,
$S_3$ denotes the 3-sphere in $\mathbb{C}^2$ : {$(x,y)\in \mathbb{C}^2 \space : \space |x|^2+|y|^2=1$}
and $\Omega=${$(x,y) \in S_3  \space | \space x^5+y^7 \neq0$}
Let $f:\Omega\rightarrow S_1$ be such that
$f(x,y)=\frac{x^5+y^7}{|x^5+y^7|}$

Prove that $f$ is a submersion.

Proof :
I will show that the differential of $f$ is a surjective map between tangent spaces.
$f$ is clearly not analytic but it should be smooth as a function of the real parts and imaginary parts of $x$ and $y$ (which are complex numbers here!).
Let $dx$ and $dy$ be «small» complex numbers (i.e. small real and imaginary parts).
$f(x+dx,y+dy)=\frac{(x+dx)^5+(y+dy)^7}{|(x+dx)^5+(y+dy)^7|}=\frac{x^5+y^7+5dx+7dy+o(dx,dy)}{|x^5+y^7|.|1+\frac{5dx+7dy}{x^5+y^7}|}$
Now, using $|z|=(z \bar z)^{1/2}$, I get that:
$$\color{red}{|1+\frac{5dx+7dy}{x^5+y^7}|^{-1}}=(1+\frac{5dx+7dy}{x^5+y^7})^{-1/2}(1+\overline{\frac{5dx+7dy}{x^5+y^7}})^{-1/2}=(1-\frac12 \frac{5dx+7dy}{x^5+y^7})(1-\frac12 \overline{\frac{5dx+7dy}{x^5+y^7}})+o(dx,dy)=1-\frac12 \frac{5dx+7dy}{x^5+y^7}-\frac12\overline{\frac{5dx+7dy}{x^5+y^7}}+o(dx,dy)=\color{red}{1-\Re(\frac{5dx+7dy}{x^5+y^7})+o(dx,dy)}$$
So,
$$\color{lime}{f(x+dx,y+dy)}=\frac{x^5+y^7+5dx+7dy+o(dx,dy)}{|x^5+y^7|}(1-\Re(\frac{5dx+7dy}{x^5+y^7})+o(dx,dy))=\color{lime}{\frac{x^5+y^7+5dx+7dy}{|x^5+y^7|}-\frac{x^5+y^7}{|x^5+y^7|}\Re(\frac{5dx+7dy}{x^5+y^7})+o(dx,dy)}$$
So identifying the linear parts, the differential of $f$ (viewed as a real function of the real and imaginary parts of $x$ and $y$, not complex!) is:
$$\color{blue}{d_{(x,y)}f(dx,dy)} = \frac{5dx+7dy}{|x^5+y^7|}-\frac{x^5+y^7}{|x^5+y^7|}\Re(\frac{5dx+7dy}{x^5+y^7})$$
$$=\frac12 \frac{5dx+7dy}{|x^5+y^7|}-\frac12 \frac{x^5+y^7}{|x^5+y^7|} \overline{\frac{5dx+7dy}{x^5+y^7}}$$
$$=\frac{1}{2|x^5+y^7|}(5dx+7dy-(x^5+y^7) \overline{\frac{5dx+7dy}{x^5+y^7}})$$
$$=\frac{x^5+y^7}{2|x^5+y^7|}(\frac{5dx+7dy}{x^5+y^7}-\overline{\frac{5dx+7dy}{x^5+y^7}})$$
$$=\frac{x^5+y^7}{2|x^5+y^7|}2i \Im(\frac{5dx+7dy}{x^5+y^7})$$
$$\color{blue}{=i\frac{x^5+y^7}{|x^5+y^7|}\Im(\frac{5dx+7dy}{x^5+y^7})}$$
$$\color{blue}{=if(x,y)\Im(\frac{5dx+7dy}{x^5+y^7})}$$
(so for this function I consider differentiability as in $\mathbb{R}^4 \rightarrow \mathbb{R}^2$ since in each of my "complex" variable I am in reality hiding two variables : the real and imaginary parts)
So now I just need to prove this constitutes a surjection from $T_{(x,y)}\Omega \rightarrow T_{f(x,y)}S_1$
We have that if $\theta$ is a regular real function then if $z(t)=e^{i\theta(t)}$, $z'(t)=i\theta'(t)e^{i\theta(t)}$ with $\theta'(t) \in \mathbb{R}$ so the tangent plane at $z$ of $S_1$ is just $T_zS_1=${${i \lambda z, \lambda \in \mathbb{R}}$}
With the same reasoning in higher dimension, $T_{(x,y)}\Omega =${$(i\lambda z, i\mu y), \lambda, \mu \in \mathbb{R}$}
When $dx, dy$ span $T_{(x,y)}\Omega$,  $df$ clearly spans $T_{f(x,y)}S_1$ since we can just identify $\lambda =\Im(\frac{5dx+7dy}{x^5+y^7})$ which should span $\mathbb{R}$ I guess as $dx, dy$ vary. So the differential is a surjection between the tangent planes and therefore $f$ is a submersion. $\square$
Does my proof look sound to you?

Going through that calculation... the differential of $x^5+y^7$ is $5x^4\,dx+7y^6\,dy$. You dropped the $x^4$ and $y^6$ factors there, in multiple places.
Now, the good news? The way you arranged your calculations, this is remarkably easy to fix. We can simply replace every instance of "$5\,dx+7\,dy$" with "$5x^4\,dx+7y^6\,dy$" and it will remain correct. The final differential is
$$df\cdot (dx,dy) = if(x,y)\operatorname{Im}\left(\frac{5x^4\,dx+7y^6\,dy}{x^5+y^7}\right)$$
On the sphere, $|f(x,y)| = 1$. In order for this differential to span the one-dimensional tangent space of $S^1$, it suffices to find some choice of $dx$ and $dy$ in the tangent space to $S^3$ so that the imaginary part there is nonzero. What is that tangent space?
$$|x|^2+|y|^2 = x\cdot \overline{x}+y\cdot \overline{y} = 1$$
$$\overline{x}\,dx+x\,\overline{dx}+\overline{y}\,dy+y\,\overline{dy} = 0$$
The tangent space is precisely the set of solutions to this linear equation, in what amounts to four variables. It's not just what you wrote down $\{(i\lambda z,i\mu y)\}$; that's the tangent space to the torus $S^1\times S^1$. Scaled copies of that $2$-manifold embed into the sphere, but there's another dimension in which $|x|$ and $|y|$ are allowed to vary.
All right, what does it take to find something in the tangent space for which the imaginary part is nonzero? Set $(dx,dy)=\left(\frac15ix,\frac17iy\right)$, so
$$\overline{x}\,dx+x\,\overline{dx}+\overline{y}\,dy+y\,\overline{dy} = \frac15i|x|^2 -\frac15i|x|^2+\frac17i|y|^2-\frac17i|y|^2=0$$
$$\frac{5x^4\,dx+7y^6\,dy}{x^5+y^7} =\frac{ix^5+iy^7}{x^5+y^7}=i$$
The first line verifies that this choice is indeed in the tangent space, and the second line verifies that the imaginary part we get out of it is nonzero. This is all valid as long as we're on the sphere and the denominator $x^5+y^7$ is nonzero - so all of $\Omega$.
