Determine if a function $f(x, y)$ is continuous in points (1, 1) and (0, 0). $$f(x, y) = \begin{cases} \frac{x^2+2xy-3y^2}{x^3-y^3}, & x \neq y \\ A, &x=y=0 \\ B, & x =y= 1\end{cases}$$
I do struggle with conclusions. How to approach it? I suppose, now one should find the iterated limits, if multivariable limit and iterated limits are equal in the point $M(x_0, y_0)$, therefore, the function is continuous in $M$ (it requires clarification, as well. I am not sure if it's viable to state so). $$1) \lim_{y\to1}\lim_{x\to1}\frac{x^2+2xy-3y^2}{x^3-y^3} = \lim_{y\to1}\frac{1+2y-3y^2}{1-y^3} = \lim_{y\to1}\frac{3(y-1)(y+\frac{1}{3})}{(y-1)(y^2+y+1)} = \frac{4}{3}$$ $$2) \lim_{x\to1}\lim_{y\to1}\frac{x^2+2xy-3y^2}{x^3-y^3} = \lim_{x\to1}\frac{x^2+2x-3}{x^3-1} = \lim_{x\to1}\frac{(x-1)(x+3)}{(x-1)(x^2+x+1)} = \frac{4}{3} $$ Proceed: $$3) \lim_{(x, y) \to (1, 1)} \frac{x^2+2xy-3y^2}{x^3-y^3} = [y = kx] = \lim_{x\to1}\frac{1+2k-3k^2}{x(1-k^3)} \Rightarrow \frac{3k^2-2k-1}{k^3-1} = \frac{3(k-1)(k+\frac{1}{3})}{(k-1)(k^2+k+1) }=[k = 2] = \frac{7}{7} = 1 \neq \frac{4}{3}$$
Now, for $A(0, 0)$: $$\lim_{(x, y)\to(0, 0)}\frac{x^2+2xy-3y^2}{x^3-y^3} = [y = 2x] = \lim_{x\to0}\frac{-7x^2}{-7x^3} = \infty$$
$f(x,y)$ is not continuous in A.
Note that when $y=x$ the numerator vanishes as well as the denominator, so you should simplify by takin a factor of $x-y$ out of both: $$ \frac{x^2+2xy-3y^2}{x^3-y^3} ={(x-y)(x+3y)\over(x-y)(x^2+xy+y^2)}$$
Now it's easy to see that $f$ is continuous at $(1,1)$ provided that $B=\frac43$ and not otherwise.
Your argument at $(0,0)$ is correct.