prove there exist one pair of functions

Question

Proof there exist only one pair of functions $z(x,y) $ and $w(x,y)$ such that $ \left\{ \begin{array}{ll} xz-yw=0 & \\ yz+xw=2 & \end{array} \right.$ and $ \left\{ \begin{array}{ll} z(1,1)=1 & \\w(1,1) =1 & \end{array} \right.$ Compute partial derivatives in $(1,1)$

Answer 1

We can't have $x=y=0$ because the second equation doesn't hold.

If $x=0$ then $yw=0$ and $yz=2$. That give us $y \ne 0$ and then: $w=0$ and $z=2/y \quad (1)$.

If $y=0$ then $xz=0$ and $xw=2$. That give us $x \ne 0$ and then: $z=0$ and $w=2/x \quad (2)$.

For $x\cdot y \ne 0$ we solve the system and get:

$$z=\frac{2y}{x^2+y^2}$$

$$w=\frac{2x}{x^2+y^2}$$

Once the above solution cover $(1)$ and $(2)$ it is the only one.