Derivatives question on partial derivatives If $z=f(x,y)$ and $x=e^u \cos v$, $y=e^u \sin v$ then show that 
$y \frac{dz}{du} + x \frac{dz}{dv} = e^{2u} \frac{dz}{dy}$
 A: First of all:
$$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}\ \mbox{ and }\ \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v},$$
where
$$\frac{\partial x}{\partial u}=e^u\cos(v)\ ,\quad \frac{\partial y}{\partial u}=e^u\sin(v)$$
$$\frac{\partial x}{\partial v}=-e^u\sin(v)\ ,\quad \frac{\partial y}{\partial v}=e^u\cos(v)$$
Such that
$$y\frac{\partial z}{\partial u}+x\frac{\partial z}{\partial v}=e^u\sin(v)\left(\frac{\partial z}{\partial x}e^u\cos(v)+\frac{\partial z}{\partial y}e^u\sin(v)\right)$$
$$+e^u\cos(v)\left(-\frac{\partial z}{\partial x}e^u\sin(v)+\frac{\partial z}{\partial y}e^u\cos(v)\right)$$
$$=e^{2u}\sin(v)\cos(v)\frac{\partial z}{\partial x}-e^{2u}\sin(v)\cos(v)\frac{\partial z}{\partial x}+e^{2u}\sin^2(v)\frac{\partial z}{\partial y}+e^{2u}\cos^2(v)\frac{\partial z}{\partial y}$$
$$=e^{2u}\bigg[\sin^2(v)+\cos^2(v)\bigg]\frac{\partial z}{\partial y}=e^{2u}\frac{\partial z}{\partial y}$$
