If the limit exists (prove) and find the value 
Let $$f(x, y) = \frac{x^2(x - 1) - y^2}{x^2 + y^2}, (x, y) \ne (0, 0)$$ Determine if $\lim_{(x, y) \to (0, 0)} f(x, y)$ exists, and if so, find the value.

Let $y = mx$, then
$$ = \lim_{x \to 0} \frac{x^2(x - 1) - m^2x^2}{x^2 + m^2x^2} = \lim_{x \to 0} \frac{x - 1 - m^2}{m^2 + 1} = -1 $$
Then we apply the squeeze theorem to prove $|f(x,y) - (-1)| < B(x)$ where $B(x) \to 0$
Main question: if we let $y = mx$, then as $y \to 0$, how can we be sure that $x \to 0$?
Since $x = y/m$ we see that if $m \ne 0$ then $x \to 0$ as $y \to 0$.
So, in the hypothesis when we say let $y = mx$, is it for non zero $m$?

Use polar coordinates: $x = r\cos\theta, \, y = r\sin\theta$ and let $r \to 0$.
EDIT
$$\begin{align}
\lim_{(x,y)\to(0,0)} \frac{x^2(x - 1) - y^2}{x^2 + y^2} 
& = \lim_{r \to 0} \frac{r^2 \cos^2\theta \, (r \cos\theta - 1) - r^2\sin^2\theta}{r^2} \\
& = \lim_{r \to 0} \left(\cos^2\theta \, (r \cos\theta - 1) - \sin^2\theta\right) \\
& = \cos^2\theta \cdot (-1) - \sin^2\theta \\
& = -1
\end{align}$$
