I'm trying to verify directly that a branch of $\sqrt{-}:\mathbb{C}\backslash(-\infty, 0) \rightarrow \mathbb{C}$ satisfies the Cauchy-Riemann equations. I've defined the branch $$ \sqrt{re^{i\theta}}:=\sqrt{r}\ \cdot e^{i\theta} $$ and verified a hint that it satisfies (for $u = \Re\sqrt{-}$, $v = \Im\sqrt{-}$): $$ u^2-v^2=x, \ 2uv = y $$
Now I guess I want to differentiate these equations, but I'm not quite sure how to go about it, not really understanding how partial derivatives work. If I differentiate $u^2-v^2=x$ with respect to $x$, etc. what is the result I get, and is this the right method?
The way you take a partial derivative is by "freezing" all the variables that you are not differentiating with respect to. By "freeze" I mean "treat as a constant". Let's jump right into your problem as an example. First I will take the partial derivative of both sides of your first equation with respect to $x$. This is easy enough, we just need to remember the chain rule for the left hand side: $$\frac{\partial}{\partial x}(u^2-v^2)=\frac{\partial}{\partial x}(x)$$ $$\implies 2u\frac{\partial u}{\partial x}-2v\frac{\partial v}{\partial x}=1$$
Now let's differentiate with respect to $y$ instead of $x$. This means that we have to treat $x$ as a constant: $$\frac{\partial}{\partial y}(u^2-v^2)=\frac{\partial}{\partial y}(x)$$ $$\implies 2u\frac{\partial u}{\partial y}-2v\frac{\partial v}{\partial y}=0$$
Now let's differentiate the second equation with respect to $x$. This means we have to treat $y$ as a constant. Also, you will need to remember the product rule: $$\frac{\partial}{\partial x}(2uv)=\frac{\partial}{\partial x}(y)$$ $$\implies 2(\frac{\partial u}{\partial x}v+u\frac{\partial v}{\partial x})=0$$
Finally, let's differentiate the second equation with respect to $y$:
$$\frac{\partial}{\partial y}(2uv)=\frac{\partial}{\partial y}(y)$$ $$\implies 2(\frac{\partial u}{\partial y}v+u\frac{\partial v}{\partial y})=1$$
Armed with these four equations, your question is asking you to verify that the Cauchy-Riemann equations hold. That is, check that $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ and $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$