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

Simpler way: Take logarithms (base e unless specified). $$m\log x+n\log y=(m+n)\log (x+y)$$ Now differentiate: $$\frac mx +\frac ny\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$$ Therefore $$\frac mx-\frac{m+n}{x+y}=\left(\frac{m+n}{x+y}-\frac ny\right)\frac{dy}{dx}$$$$\implies \frac{my-nx}{x(x+y)}= \frac{my-nx}{y(x+y)} \cdot \frac{dy}{dx}$$ so $$\frac{dy}{dx}=\frac yx.$$
