Consider the function,
$f(x,y)=\dfrac{x^3\cos\left(\frac1y\right)+y^3\cos\left(\frac1x\right)}{ x^2+y^2},\qquad\forall x,y\neq0$
And $f(x,y)= 0,\qquad\text{ if } xy=0$
Clearly this function is not continuous at either $x-$ or $y-$ axis
As, when approaching $x-$axis,
$f(x,y)=\dfrac{x^3\cos\left(\frac1y\right)}{ x^2} = x\cos\left(\dfrac1y\right) \neq 0$
But how do you tell at origin. Approaching O via path $y=mx$ does not prove that it is discontinuous.
Best Answer
By inequalities and polar coordinates we have that
$$0\le \left|\frac{x^3\cos(1/y)+y^3\cos(1/x)}{ x^2+y^2}\right|\le \left|\frac{x^3\cos(1/y)}{ x^2+y^2}\right|+\left|\frac{y^3\cos(1/x)}{ x^2+y^2}\right|\le$$$$\le \left|\frac{x^3}{ x^2+y^2}\right|+\left|\frac{y^3}{ x^2+y^2}\right|=r\left(|\cos^3\theta|+|\sin^3\theta|\right)\to 0$$
For the limit when approching the $x$ axis at a point $x=a\neq 0$ we have that
$$\lim_{y\to 0} \frac{a^3\cos(1/y)+y^3\cos(1/a)}{ a^2+y^2}=\lim_{y\to 0} \frac{a^3\cos(1/y)}{ a^2+y^2}+\lim_{y\to 0} \frac{y^3\cos(1/a)}{ a^2+y^2}$$
and the second limit is equal to $0$ but the first one doesn't exist indeed for
$y_n=\frac1{2\pi n}\to 0 \implies \frac{a^3\cos(2\pi n)}{ a^2+\frac1{4\pi^2 n^2}}\to a$
$y_n=\frac1{(2n-1)\pi}\to 0 \implies \frac{a^3\cos((2n-1)\pi)}{ a^2+\frac1{(2n-1)^2\pi^2}}\to -a$