Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a function defined by

$f(x)=\{\begin{array}{cc}{x}^2\sin <!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{x^2}\right),& \text{ if }x\ne <!--\; \ne \; -->0\\ 0,& \text{ if }x=0\end{array}$

Then which of the following statements is TRUE?

The given function is defined as:
$f(x)=x^2\sin \left(\frac{\pi }{x^2}\right)$
for all $x\ne 0$, and $f(0)=0$.

We want to find the number of solutions to the equation $f(x)=0$ in the interval $(\frac{1}{{\pi }^2},\frac{1}{\pi })$.

For any $x\ne 0$, the equation $f(x)=0$ simplifies to:
$x^2\sin \left(\frac{\pi }{x^2}\right)=0$
Since $x\ne 0$, we must have:
$\sin \left(\frac{\pi }{x^2}\right)=0$

The sine function vanishes at integer multiples of $\pi$. Therefore:
$\frac{\pi }{x^2}=n\pi$
for some integer $n\ne 0$. Simplifying this expression, we get:
$x^2=\frac{1}{n}$

Since we are looking for solutions in the positive interval $(\frac{1}{{\pi }^2},\frac{1}{\pi })$, $x$ must be positive. Thus, we have:
$x=\frac{1}{\sqrt{n}}$
where $n$ is a positive integer.

Now we apply the interval constraint:
$\frac{1}{{\pi }^2}<\frac{1}{\sqrt{n}}<\frac{1}{\pi }$

Taking the reciprocal of all terms reverses the inequalities:
$\pi <\sqrt{n}<{\pi }^2$

Squaring all parts of the inequality gives:
${\pi }^2<n<{\pi }^4$

Let us approximate the numerical values of the boundaries:
$\pi \approx 3.14159$
${\pi }^2\approx 9.8696$
${\pi }^4={\left({\pi }^2\right)}^2\approx {9.8696}^2\approx 97.409$

Since $n$ must be an integer, it must satisfy:
$9.8696<n<97.409$
Thus, the possible integer values for $n$ are:
$n\in \{10,11,12,...,97\}$

The total number of such integers is:
$97-10+1=88$

Since there are exactly 88 solutions, the statement that $f(x)=0$ has more than 25 solutions in the interval $(\frac{1}{{\pi }^2},\frac{1}{\pi })$ is indeed TRUE.