The tangent to the curve  $y=x{e}^{x^2}$  at the point  $(1,e)$  also passes through the point

$\begin{array}{l}y=x{e}^{x^2}\\ \frac{dy}{dx}=e^{x^2}+x.e^{x^2}\times 2x=e^{x^2}(2x^2+1)\\ slopem=\frac{dy}{dx}at(1,e)\\ =3e\\ Equationoftangent\\ y-e=3e(x-1)\\ 3ex-y-2e=0\\ (\frac{4}{3},2e)liesonaboveline\end{array}$