For the limit to exist it must be the same no matter which direction we approach it. I'd recommend approaching the origin along the lines y = x and y = -x.
y = x:
$\displaystyle \displaystyle \lim_{(x,y) \to (0,0)} \frac{sin(x^2)~sin^3(y)}{x^6 + y^4}$
$\displaystyle \displaystyle = \lim_{x \to 0} \frac{sin(x^2)~sin^3(x)}{x^6 + x^4}$
This is in the form to use l'Hopital's rule, but its a mess to do it this way. Expanding a bit:
$\displaystyle \displaystyle = \lim_{x \to 0} \left ( \frac{sin(x^2)}{x^2} \cdot \frac{sin(x)}{x} \cdot \frac{sin(x)}{x} \cdot \frac{sin(x)}{x^2 + 1} \right )$
$\displaystyle \displaystyle = \lim_{x \to 0} \frac{sin(x^2)}{x^2} \cdot \lim_{x \to 0} \frac{sin(x)}{x} \cdot \lim_{x \to 0} \frac{sin(x)}{x} \cdot \lim_{x \to 0} \frac{sin(x)}{x^2 + 1}$
$\displaystyle \displaystyle = 1 \cdot 1 \cdot 1 \cdot \lim_{x \to 0} \frac{sin(x)}{x^2 + 1}$
This last limit is 0. y = -x is done in the same way.
-Dan