How can the limit of this $F(x,y)$ is $0$ when it is taken along $y=x^2$ or $ x=y^2$? 
Find $$\lim_{(x,y) \to (0,0)}\frac{3x^{2}y}{x^{2}+y^{2}}$$ if it
  exists.

From my textbook, it said that the limits along the parabolas $$y=x^{2}\text{ and } x=y^{2}$$ also turn out to be 0. 
I couldn't figure out why, as if we set $$x=y^{2}$$, for example
then $$\lim_{(x,y) \to (0,0)}\frac{3x^{4}}{x^{2}+(x^{2})^{2}}$$
$$\lim_{(x,y) \to (0,0)}\frac{3x^{4}}{x^{2}+x^{4}}$$
$$\lim_{(x,y) \to (0,0)}\frac{3}{0+1}$$
then the answer is supposed to be $3$?  right?

On $y=x^2$ we have
$$
\frac{3x^2y}{x^2+y^2}=\frac{3x^4}{x^2+x^4}=
\frac{3x^4}{x^2(1+x^2)}=\frac{3x^2}{1+x^2}
$$
which has limit $0$ when $x\to0$.
On $x=y^2$ we have
$$
\frac{3x^2y}{x^2+y^2}=\frac{3y^5}{y^4+y^2}=\frac{3y^5}{y^2(y^2+1)}
=\frac{3y^3}{y^2+1}
$$
which again has limit $0$ for $y\to0$.
