3-Variable Limit $\lim\limits_{(x,y,z,) \to (0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$ $$\lim_{(x,y,z,) \to (0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$$
Background: I think that converting the formula into parametric variables won't work since I only know that's useful for converting $\sin^2(x) + \cos^2(x)$ into $1$. I've actually never done a limit with three variables before so I'm a little confused about how we can go about it. Any suggestions as to how I should begin to approach this?

Approach along $x=y=z=a$:
$$\lim_{(x,y,z)\to(0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}=\lim_{a\to0}\frac{a^2+2a^3}{2a^2+a^4}=\lim_{a\to0}\frac{1+2a}{2+a^2}=\frac12$$
Approch along $x=z=a$, $y=a^2$:
$$\lim_{(x,y,z)\to(0,0,0)} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}=\lim_{a\to0}\frac{2a^3+a^4}{a^2+2a^4}=\lim_{a\to0}\frac{2a+a^2}{1+2a^2}=0$$
Hence, the limit doesn't exist.
