Evaluate the following limit:
\[
\lim _{x \rightarrow 0} x^{3} \sin \left(\frac{1}{x^{2}}\right)=
\]

Step 1 ：First, we observe that the function \(x^{3} \sin \left(\frac{1}{x^{2}}\right)\) is undefined at \(x=0\). However, we can still evaluate the limit as \(x\) approaches \(0\).

Step 2 ：We can rewrite the function as \(x \cdot x^{2} \sin \left(\frac{1}{x^{2}}\right)\).

Step 3 ：Next, we apply the squeeze theorem. We know that \(-1 \leq \sin \left(\frac{1}{x^{2}}\right) \leq 1\) for all \(x\), so \(-x^{2} \leq x^{2} \sin \left(\frac{1}{x^{2}}\right) \leq x^{2}\).

Step 4 ：Multiplying through by \(x\), we get \(-x^{3} \leq x^{3} \sin \left(\frac{1}{x^{2}}\right) \leq x^{3}\).

Step 5 ：As \(x\) approaches \(0\), both \(-x^{3}\) and \(x^{3}\) approach \(0\). Therefore, by the squeeze theorem, \(\lim _{x \rightarrow 0} x^{3} \sin \left(\frac{1}{x^{2}}\right)\) must also be \(0\).

Step 6 ：So, \(\lim _{x \rightarrow 0} x^{3} \sin \left(\frac{1}{x^{2}}\right) = \boxed{0}\).