I'm a bit confused about the multivariable differentiability criterion and what you're supposed to check.
Let me see if I have this correct. The theorem is:
If the partials EXIST AND ARE CONTINUOUS at a point, then the function is differentiable at that point.
So:
-To check the EXISTENCE of a partial derivative at a point, you take the 1-Dimensional limit as you approach the point along a single path of the single variable coordinate (i.e., substitute the constant point values for all other variables).
-To check the CONTINUITY of a partial derivative at a point, you take the n-Dimensional limit as you approach the point along any path. (i.e., compute the partial formula as usual by treating all other variables as general constants, but then allow them all to vary when taking the limit along any path)
In other words, the confusion is that by saying "the partials exist and are continuous" it makes it sound like you're checking 1-D existence and 1-D continuity, but that's wrong. It's 1-D existence but n-D continuity. (Is that correct?)
For example, we want to investigate the differentiability at (0,0) of:
$$f(x,y)= \begin{cases} \frac{x^2y}{x^2+y^2}&(x,y)\neq(0,0)\\ 0&(x,y)=(0,0)\\ \end{cases} $$
The partials EXIST, as follows (only x-case, since y-case is similar):
$$f_x(0,0)=\lim_{h\to0} \frac{f(h,0)-f(0,0)}{h}= \lim_{h\to0}\frac{0}{h^3}=0$$
But to check CONTINUITY, you have to take:
$$f_x(x,y)= \frac{2xy^3}{(x^2+y^2)^2}, (x,y)\neq (0,0)$$
and check (only x-case, but you'd have to check that all work):
$$\lim_{(x,y)\to(0,0)} \frac{2xy^3}{(x^2+y^2)^2}$$
which doesn't exist, so we can't apply the theorem.
Is that the correct understanding of what needs to be done?