Let $f(x,y) = x^3 + y^3 -3x$. Since it's everywhere differentiable, we know that its only critical points are $(1,0),(-1,0)$ given that the gradient $\nabla f(x,y) = (3x^2-3,3y^2)$ is zero if and only $x=\pm1$ and $y=0$.
However, $\det Hf|_{(\pm1,0)} = 0$, so we can't apply the Second partial derivative test.
Since $f$ is odd, we might suspect these are saddle points. One way to proceed is to find curves:
$$f \circ \alpha (t) = f\circ (t^2+1, (3^{\frac{1}{3}}t^{\frac{2}{3}})) = (t^2+1)^3 -3$$
$$f \circ \beta (t) = f\circ (-t^2+1, (-3^{\frac{1}{3}}t^{\frac{2}{3}})) = (-t^2+1)^3 -3$$
And it's clear that $f \circ \alpha (t)$ has a minimum at $t=0$ and that $f \circ \beta (t)$ has a maximum at $t=0$, which implies $(1,0)$ is a saddle point of $f$.
But this method depends on one's ingeniousness and it's very time-consuming. I've lost half an hour trying to find those curves!
Are there any other strategies to tackle these kinds of problems?