Solution: If we denote \[ f(x,y) = \frac{y\sin \left( \frac{x}{\sqrt{y} } \right) }{\sqrt{x^2 + y^2} }, \] then we note that $f(x,y) = -f(-x,y)$. So it is sufficient to find the limit for $x\geq 0$. Recall that for $\theta \geq 0$, \[ \sin \theta \leq \theta . \] So, for $x\geq 0$, we have \begin{align*} \sin \left( \frac{x}{\sqrt{y} } \right) \leq \frac{x}{\sqrt{y} }. \end{align*}

Since $y > 0$, we have \[ y\cdot\sin \left( \frac{x}{\sqrt{y} } \right) \leq y\cdot\frac{x}{\sqrt{y} } = x\sqrt{y}. \] Therefore, \begin{align*} \frac{y\sin \left( \frac{x}{\sqrt{y} } \right) }{\sqrt{x^2 + y^2} } \leq \frac{\sqrt{y} x}{\sqrt{x^2 +y^2} } \leq \frac{\sqrt{y} x}{\sqrt{x^2} } = \sqrt{y}. \end{align*} So we have \[ 0 \leq \frac{y\sin \left( \frac{x}{\sqrt{y} } \right) }{\sqrt{x^2 + y^2} } \leq \sqrt{y}. \] Applying the Sandwich's lemma to obtain the limit of $f$ \[ \lim_{(x,y) \to (0,0)} \frac{y\sin \left( \frac{x}{\sqrt{y} } \right) }{\sqrt{x^2 + y^2} } = 0. \]