Find $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial v}$ if $z=f(x,y)$ , $u=e^x\cos y$ , $v=e^{-x}\sin y$

Question

I don't know how to use the fact that $z=f(x,y)$.

I tried to solve $(x,y)$ in terms of $(u,v)$ and found that $y=\frac{\sin^{-1}(2uv)}{2}$ but can't solve $x$.

Another question: Is $\frac{\partial x}{\partial u}=\frac{1}{\frac{\partial u}{\partial x}}$ ?

Answer 1

If you take $u=e^x \cos y$ and differentiate with respect to $u$, you get $$ 1 = e^{x} \cos y \frac{\partial x}{\partial u} - e^x \sin y \frac{\partial y}{\partial u} $$ If you take $v = e^{-x}\sin y$ and differentiate with respect to $u$, you get $$ 0 = -e^{-x} \cos y \frac{\partial x}{\partial u} - e^{-x} \sin y \frac{\partial y}{\partial u} $$ This is a system of two equations for the two unknowns $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial u}$.

Can you solve it from here?