Implicit Differentiation - $x^my^n = (x+y)^{m+n}$ 
Use implicit differentiation to find $\frac{\mbox{dy}}{\mbox{dx}}$ if $$x^my^n = (x+y)^{m+n}$$

Differentiation both sides with respect to $x$:
$$mx^{m-1}y^n + x^mny^{n-1}y' =(m+n)(x+y)^{m+n-1}(1 + y')$$
$$y' = \frac{(m+n)(x+y)^{m+n-1} - mx^{m-1}y^n}{x^mny^{n-1}-(m+n)(x+y)^{m+n-1}}$$
$$y'= \frac{nxy-my^2}{nx^2-mxy}$$ after using given $x^my^n = (x+y)^{m+n}$. The answer given to me is $\frac{y}{x}$. So, it seems $y'$ can be simplified even more.
How do we do this?
Thanks

For any homogeneous relation of $x$ and $y$, we always have $$\frac{dy}{dx}=\frac{y}{x}.$$ Because the relationship is always satisfied by $y=vx$ meaning $$\frac{dy}{dx}=v=\frac{y}{x}.$$ The relation $$x^my^n=(x+y)^{m+n}$$ is a homogeneous function of degree $m+n$.
For $x^2+xy+y^2=0$ and $4x^2y+5y^3+6x^3-7xy^2=0.$
After finding you have to use back the relationship to get
$\frac{dy}{dx}=\frac{y}{x}$
