The tangent to the curve,

The equation of curve $y=x e^{x^{2}}$

$\Rightarrow \quad \frac{d y}{d x}=e^{x^{2}} \cdot 1+x, \mathrm{e}^{x^{2}} \cdot 2 x$

Since $(1, e)$ lies on the curve $y=x e^{x^{2}}$, then equation of tangent at $(1, e)$ is

$y-e=\left(e^{x^{2}}\left(1+2 x^{2}\right)\right)_{x=1}(x-1)$

$y-e=3 e(x-1)$

$3 e x-y=2 e$

So, equation of tangent to the curve passes through the

point $\left(\frac{4}{3}, 2 e\right)$