If $\mathrm{f}(x)=\frac{\sin \left(\pi \cos ^2 x\right)}{3 x^2}, x \neq 0$ is continuous at $x=0$ then $\mathrm{f}(0)=$

To ensure continuity at $x = 0$, we require $\mathrm{f}(0) = \lim_{x \to 0} \mathrm{f}(x)$, where $\mathrm{f}(x) = \frac{\sin \left(\pi \cos^2 x\right)}{3 x^2}$ for $x \neq 0$.

Rewriting using $\cos^2 x = 1 - \sin^2 x$, the argument becomes $\pi \cos^2 x = \pi (1 - \sin^2 x) = \pi - \pi \sin^2 x$. Using the identity $\sin(\pi - \theta) = \sin \theta$, we have $\sin(\pi \cos^2 x) = \sin(\pi \sin^2 x)$.

The limit evaluates as:

$\lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{3 x^2} = \lim_{x \to 0} \left[ \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{3 x^2} \right]$

Applying standard limits: as $x \to 0$, $\pi \sin^2 x \to 0$, so the first factor tends to 1.
The second factor simplifies to $\frac{\pi}{3} \left( \frac{\sin x}{x} \right)^2 \to \frac{\pi}{3}$.

Therefore, $\mathrm{f}(0) = \frac{\pi}{3}$.

$\boxed{\frac{\pi}{3}}$