I'm having some troubles in understanding how to proceed on this exercise. The requests are: does it exist $L \in\mathbb{R}$ such that $f(x, y)$ is continuous at $(-1, 0)$? Which partial derivatives among $f'_x(-1, 0)$ and $f'_y(-1, 0)$ do exist for all values of $L$ and why?
$$f(x, y) = \begin{cases} \dfrac{x^2+x}{x^2-y^2-1}, & x^2 - y^2 \neq 1 \\\\ L, & x^2-y^2 = 1\end{cases}$$
ATTEMPTS
The first request "seems" easy, unless I made a mess. I thought the limits must be coherent, namely equal and finite, that is:
$$\lim_{x\to -1} \left(\lim_{y \to 0} \dfrac{x^2+x}{x^2-y^2-1}\right) = \lim_{y\to 0}\left(\lim_{x\to -1} \dfrac{x^2+x}{x^2-y^2-1}\right)$$
Which doesn't occur because the lhs limit evaluates to $1/2$ whereas the rhs limit evaluates to $0$.
Now my problem is to understand if the function is continuous for some values of $L$, because for the partial derivative request I thought to use the difference quotient, hence the definition of the derivative. But when I do it, the $L$ remains there.
I cannot use polar coordinates, by the way (if this makes sense to be told).
Some work
So going with the definition of the derivative, evaluating ad $(-1, 0)$ I get:
$$f'_x(-1, 0) = \lim_{h\to 0} \frac{\frac{(-1+h)^2 - (-1+h)}{(-1+h)^2 - 1} - L}{h}$$
This turns out to be $$f'_x(-1, 0) = \lim_{h\to 0} \frac{1}{h}\left(1 - \frac{1}{h} - L\right)$$
Hence the limit is infinite and $f'_x(-1, 0)$ does not exist.
Then
$$f'_y(-1, 0) = \lim_{h\to 0} \frac{\frac{-1+1}{1-h-1} - L}{h}$$
But this turns out to be infinite too...