Let $$ \begin{align} u = \begin{cases} \sqrt{r} \cosh(t) & \text{if } r\gt 0\\ \sqrt{-r} \sinh(t) & \text{if } r \lt 0\\ \end{cases} \end{align} \tag 1 $$ and $$ \begin{align} v = \begin{cases} \sqrt{r} \sinh(t) & \text{if } r\gt 0\\ \sqrt{-r} \cosh(t) & \text{if } r \lt 0\\ \end{cases} \end{align} \tag 2 $$
Combining both $(1)$ and $(2)$ and using $\cosh^2-\sinh^2=1$, $ u^2 - v^2=r \tag 3$
Using $(1)$, $$\frac{\partial{u}}{\partial{r}}=\frac{\cosh(t)}{2\sqrt{r}} \text{ if } r\gt 0 \quad \lor \quad -\frac{\sinh(t)}{2\sqrt{-r}} \text{ if } r\lt 0$$
However, using $(3)$,
$$ \frac{\partial{u}}{\partial{r}}= \frac{1}{2u} = \frac{1}{2\sqrt{r}\cosh(t)}\text{ if } r\gt 0 \quad \lor \quad \frac{1}{2\sqrt{-r}\sinh(t)}\text{ if } r\lt 0$$
Why don't both of the derivatives of $\frac{\partial{u}}{\partial{r}}$ match? What did I do wrong?
You cannot use partial derivatives when implicitly differentiating; you must use the total derivative. Otherwise you "miss" the fact that $v$ is not independent of $r$: $$2u \frac{du}{dr} - 2v \frac{dv}{dr} = 1.$$