Changing variables for multivariable functions 
Let $x=uv$ and $y=\frac{1}{2}(u^2-v^2)$. Substitute $x$ and $y$ with $u$ and $v$ in the expression $z_{x}^2+z_y^2$

My attempt:
$$z_x=u_xz_u+v_xz_v=\frac{1}{v}z_u+\frac{1}{u}z_v$$
Squaring:
$$z_x^2=\frac{1}{v^2}z_u^2+\frac{2}{vu}z_uz_v+\frac{1}{u^2}z_v^2$$
Using the same method for $z_y$
$$z_y=\frac{1}{\sqrt{2y+v^2}}z_u-\frac{1}{\sqrt{u^2-2y}}z_v$$
Squaring again and changing variables and summing:
$$z_x^2+z_y^2=\frac{1}{v^2}z_u^2+\frac{2}{vu}z_uz_v+\frac{1}{u^2}z_v^2+\frac{1}{u^2}z_u^2-\frac{2}{uv}z_uz_v+\frac{1}{v^2}z_v^2=(\frac{1}{v^2}+\frac{1}{u^2})(z_u^2+z_v^2)$$
My book gives the answer $\frac{z_u^2+z_v^2}{u^2+v^2}$. I checked my calculations multiple times and there doesn't seem to be any errors so I assume my reasoning was incorrect somewhere. Any help?

Tackle it from the other side, express $z_u^2 + z_v^2$ in terms of $z_x,\,z_y$. You have
$$z_u = z_x x_u + z_y y_u = vz_x + uz_y,$$ and $$z_v = z_xx_v + z_y y_v = uz_x - vz_y.$$ That produces $z_u^2 + z_v^2 = (u^2+v^2)(z_x^2+z_y^2)$, agreeing with the book.
What was your mistake? When computing $u_x$ etc., you wrote $u = \frac{x}{v}$ to conclude $u_x = \frac{1}{v}$, as though $v$ was independent of $x$. But it isn't, $v$ is a function of $x$ and $y$, like $u$ is. You would get
$$u_x = \left(\frac{x}{v}\right)_x = \frac{1}{v} - \frac{xv_x}{v^2}$$
correctly from the ansatz $u = \frac{x}{v}$. That doesn't look too nice, the hindsight computation above is much nicer, but of course one can only do that if one knows what one will obtain. Although one knows it is some quadratic combination of $z_u$ and $z_v$, and one only needs to determine the coefficients, which isn't too bad, since there are only $z_u^2,\, z_uz_v,\,z_v^2$.
To compute $u_x,\, u_y,\, v_x,\,v_y$, it is maybe the easiest to invert the matrix
$$\begin{pmatrix}x_u & x_v \\ y_u & y_v \end{pmatrix} = \begin{pmatrix} v & u\\ u & -v \end{pmatrix}$$
and get
$$\begin{pmatrix}u_x & u_y \\ v_x & v_y \end{pmatrix} = \frac{1}{u^2+v^2}\begin{pmatrix} v & u\\ u & -v \end{pmatrix}.$$
