Let $$f(x,y)= \begin{cases}\dfrac{xy^3}{x^3+y^6}& \text{if } x^3+y^6\not =0\\ \\ 0&\text{if } x^3+y^6=0 \end{cases} $$
Is $f$ continuous?
If $x=y^2$ then we have:
$$\lim_{y\rightarrow 0}\frac{y^2y^3}{y^6+y^6}=\lim_{y\rightarrow 0}\frac{y^5}{2y^6}=\frac{1}{2}\lim_{y\rightarrow0} \frac{1}{y}$$ And that limit doesn't exist. Then $f$ is not continuous. I'm not sure of the process, can someone review my proof?
Alternatively, you could consider
$$f\left(|t|,\sqrt{|t|}\right) = \frac{|t|^{5/2}}{|t|^3+|t|^3} = \frac{t^5}{2t^6} = \frac{1}{2\sqrt{|t|}} \to \infty$$ Clearly $$\lim_{t\to0}f(t^2,t)\neq 0$$
Then $f$ is not continuous. or $$f(t^2,|t|) = \frac{|t|^5}{2t^6} = \frac{1}{2|t|}$$