When can L'Hospital rule be used on Multivariable limits. I am wondering about a multivariable limit, and in particular, is it ever valid to use L'hospital rule.
For example, I am working on $$ \lim_{(x,y) \to (1,1)} \frac{x^3-y}{x-y}$$
This is what I have done,
let $$f(x,y)=\frac{x^3-y}{x-y}$$
$f(x,0) \rightarrow 1$ as $(x,y) \rightarrow (1,1)$
and similiary
$f(0,y) \rightarrow 1$ as $(x,y) \rightarrow (1,1)$
Okay now here is where I have a few questions ( I haven't looked at the answer or used wolfram or anything because I want to make sure I understand it first), should I continue to try out different parts, or should I try to see if I can prove the limit is 1.
in trying different paths, say $$f(x,x^2)=\frac{x^2(1-x^3)}{(1-x)}$$ would it now be valid to use L'hospital? because the y is gone and we would have 0/0 as x $\rightarrow 1$? or is it never valid to use this rule for multi valued?
Is this the right approach I should be taking or is there something else I should be thinking of?
Thank you
 A: Bear in mind the L'Hospital's rule goes for single-variable limits, only. Checking a lot of different paths will not guarantee the existence of the limit. But if you find any two different paths which give you different numbers, then the limit does not exists.
That being said, once you have chosen a path, the limit becomes a single-variable on, so yes, you can use L'Hospital. For example:

in trying different paths, say $f(x,x^2)=\frac{x^2(1−x^3)}{(1−x)}$ would it now
  be valid to use L'hospital? because the $y$ is gone and we would have
  $0/0$ as $x →1$?

Here you chose a path, and now you have a single-variable limit. You can use L'Hospital.

Edit: It seems that there is a sort of L'Hospital's rule for multi-variable limits, as pointed by Git Gud in the comments. Check it out.
A: There will be some cases where L'Hopital's rule naively applied will give the right answer for a double limit.  But to understand when you are in such a case would require a deeper knowledge of the function, so l'Hopital's rule is pretty useless for proving the limit exists (although it is still useful to prove tha a limit does not exist).
A: Set $x=1+t, y=1+t^{\alpha}$. Then
$$f(x,y)=\frac{t^3+3t^2+3t-t^{\alpha}}{t-t^{\alpha}}\sim_0\begin{cases}\dfrac{3t}t=3&\text{if}\enspace \alpha>1,\\\dfrac{-t^{\alpha}}{t^{\alpha}}=-1&\text{if}\enspace 0<\alpha<1\end{cases}.$$
Hence the limit is $3$ if $\alpha>1$, $-1$ if $0<\alpha<1$. This proves there is no limit at $(1,1)$.
A: enter link description hereL.Hopital rule is used in the case of indeterminate forms. the present example is suitable for existence limits at $(1,1)$ but not equal.
This way, limit does not exist is the conclusion. 
Therefore, this example is not suitable for L.Hopital rule for multivariate function. 
