I need help in this exercise. What I need to prove that the function $f$ given is not continuous in the point $(0,0)$
$$ f(x,y) = \begin {cases} \frac {x^3\times y} {x^6+y^2} & (x,y) \not = 0\\ 0 & (x,y) = 0 \end {cases} $$
So what I've done so far is to calculate the limit of the function in the first place with two variables:
$$ \lim_{(x,y)\to\ (0,0)} \frac {x^3\times y} {x^6+y^2} $$ I substitute $y=mx$ which is the slope $$ \lim_{(x)\to\ (0)} \frac {x^3\times mx} {x^6+(mx)^2} $$ $$=\lim_{(x)\to\ (0)} \frac {x^4\times m} {x^6+m^2x^2} $$ $$=\lim_{(x)\to\ (0)} \frac {x^4\times m} {x^2(x^6+m^2)} $$ $$=\lim_{(x)\to\ (0)} \frac {x^2\times m} {x^6+m^2} $$ $$=\lim_{(x)\to\ (0)} \frac {x^2\times m} {x^6+m^2} $$ $$=\frac {0^2\times m} {0^6+m^2} = 0$$
So my result says that it is continuous. What have I done wrong? What do I need to do to prove that it is not if I already calculated that it is? Thank you so much. If something isn't very clear, please let me know.
We consider the path $y=x^3$ to $(0,0)$. Along this path, the function becomes $$\frac{x^3\cdot x^3}{x^6+x^6}=\frac{x^6}{2x^6}=\frac12$$ and so the limit along this path is $\frac12$. Since this is different from the limit of 0 you obtained with the different path $y=mx$, the limit at the origin does not exist.