Question: I have to find $u(x,y)$ and $v(x,y)$ such that $f=u+iv$ is analytic. The textbook says that: $$2xyu+(y^2-x^2)v+2xy(x^2+y^2)^2=0$$

I've already tried to divide all the relation by $2xy$ (supposing $x$, $y$ non-zero) and then calculating the second order partial derivatives (to prove that $u$ or $v$ are harmonic). The computations got pretty complicated, so I was wondering, is there an easier way to find $u$ and $v$?

Answer: 
No.

It's better:
$$(2^x)^2+2^x-6=0$$ or
$$(2^x-2)(2^x+3)=0$$ or $$2^x=2,$$ which gives $$x=1.$$
Also, you can say that $2^{2x}+2^x$ increases, which by your work gives $x=1$ again.
You should be careful because your "rule" does not always hold.

It is always true that $x = y \implies b^x = b^y$ because the function $f\_b(x) = b^x$ is well-defined (assume we are working with positive real numbers). 

It is not always the case of the converse, $b^x = b^y \implies x = y$. When $f(x) = f(y) \implies x = y$, the function is *injective*. 

Observe for example the function $f\_1(x) = 1^x$ is *not injective* since $1^1 = 1^2$ but $1 \ne 2$. However, for $b \ne \pm 1$ (ignoring the case $b=0$), $f\_b(x) = b^x$ is injective (try to prove this).