Continuity of multivariate functions Determine whether the function
$$
f(x,y)=\left\{\begin{array}{@{}l@{}}
(x^3-y^3)/(x^2+y^2) & \text{for }(x,y)\neq(0,0)\\
0 & \text{for }(x,y)=(0,0)\end{array}\right.
$$
is continuous.
To do so, I test whether the limit of the function $f(x,y)=\frac{x^3-y^3}{x^2+y^2}$ exists by plugging $y=mx$. If the limit gives the same value for all $m$, it means that the function exists and hence is continuous. I'm unsure whether my method is correct.
 A: By the inequality
$$0\le\left|\frac{x^3-y^3}{x^2+y^2}\right|\le\frac{|x|x^2+|y|y^2}{x^2+y^2}\le|x|+|y|\xrightarrow{(x,y)\to(0,0)}0$$
we deduce the continuity on $0$.
A: In general, the method you are using can only disprove continuity if you find distinct limit values, i.e. a limit depending from $m$. However, it will not prove the limit exists, as you could still find a different limit approaching from, say, $y=x^2$. One way to determine a limit is by the polar coordinates. Substitute:
$$\left\{\begin{array}{@{}l@{}}
x=\rho\cos\theta \\
y=\rho\sin\theta
\end{array}\right.,$$
or, more in general, $x-x_0$ and $y-y_0$ instead of $x$ and $y$, when calculating a limit for $(x,y)\to(x_0,y_0)$. In our case, we have:
$$\lim_{(x,y)\to(0,0)}\frac{x^3-y^3}{x^2+y^2}=\lim_{\rho\to0^+}\frac{\rho^3(\cos^3\theta-\sin^3\theta)}{\rho^2(\cos^2\theta+\sin^2\theta)}=\lim_{\rho\to0^+}\rho\cdot(\cos^3\theta-\sin^3\theta),$$
which is naturally 0 since it is $\rho$, which tends to 0, times a limited quantity. Hence the function is continuous.

How do u know which method to use?

Well, you can try the polar coordinates. If that doesn't help, you can try using inequalities. In fact, if you see a useful inequality, you can try using it from the start, as that will probably spare you some calculations. If none of that helps, then you can try your method because at that point you will probably have reasons to believe the limit depends on the curve you are approaching from, i.e. does not exist.
A: No it's not correct generally.
To prove that the function $f$ is continuous at $(x_0,y_0)$ you need to prove that (which is, as I know, the definition of continuity):
$\forall \varepsilon>0:\exists\delta>0$ such that:
$\forall x,y$ satisfying$\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ we have $|f(x,y)-f(x_0,y_0)|<\varepsilon$
In the answer of Sami, he/she basically need to find $\delta$ so that $|x|+|y|<\varepsilon$ to satisfy the conditions, which is easy by putting $\delta=\frac{\varepsilon
}{2}$
