The number of critical points for the function $f\left(x\right)=x{e}^{x^2}$ are.

Given: $f\left(x\right)=x{e}^{x^2}$

Critical points are the points where either $f\text{'}\left(x\right)=0$ or $f\text{'}\left(x\right)$ is not defined.

Step-$1$ Differentiating, $f\left(x\right)$ with respect to $x$ we get ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\left(1\right)·{\mathrm{e}}^{{\mathrm{x}}^2}+2\mathrm{x}\left({\mathrm{e}}^{{\mathrm{x}}^2}\right)\left(\mathrm{x}\right)$,

$$\begin{gathered} \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)={\mathrm{e}}^{{\mathrm{x}}^2}+2{\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^2} \\ \Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)={\mathrm{e}}^{{\mathrm{x}}^2}\left(1+2{\mathrm{x}}^2\right). \end{gathered}$$

Step-$2$ Put, $f^{\text{'}}\left(x\right)=0$ To calculate critical points $.$

$\Rightarrow {\mathrm{e}}^{x^2}+2x^2{e}^{x^2}=0$

$\Rightarrow {\mathrm{e}}^{{\mathrm{x}}^2}\left(1+2{\mathrm{x}}^2\right)=0$

Since, $1+2x^2>0\forall x\in R$

Also , ${\mathrm{e}}^{x^2}>0\forall x\in R$

It is clear that always $f^{\text{'}}\left(x\right)>0$, Hence their will be no critical point in $f\left(x\right)=x{e}^{x^2}$.

Hence option-$4$ is the correct answer.