Answer
Does not exist
Work Step by Step
Evaluate the limit along $x=t$, $y=t$, $z=t$:
$\lim\limits_{(x,y,z) \to (0,0,0)}\frac{yz}{x^2+4y^2+9z^2}=\lim\limits_{t \to 0}\frac{t\cdot t}{t^2+4t^2+9t^2}=\lim\limits_{t \to 0}\frac{t^2}{14t^2}=\lim\limits_{t \to 0}\frac{1}{14}=\frac{1}{14}$
Now, evaluate the limit along $x=t$, $y=t$, $z=-t$:
$\lim\limits_{(x,y,z) \to (0,0,0)}\frac{yz}{x^2+4y^2+9z^2}=\lim\limits_{t \to 0}\frac{t\cdot (-t)}{t^2+4t^2+9(-t)^2}=\lim\limits_{t \to 0}\frac{-t^2}{14t^2}=\lim\limits_{t \to 0}\frac{-1}{14}=-\frac{1}{14}$
Since the limits along the two curves which lie the origin are different, the value of the limit does not exist.
