If $f\left(x\right)=3e^{x^2},$ then $f^1\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^1\left(0\right)=$

We have $f\left(x\right)=3e^{x^2},$ Differentiating w.r.t. x, we get $f^{\mathrm{\prime }}\left(x\right)=6x{e}^{x^2}$; 

$\therefore f\left(0\right)=3 \text{and }{f}^{\mathrm{\prime }}\left(0\right)=0$

$\Rightarrow f^{\mathrm{\prime }}\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^{\mathrm{\prime }}\left(0\right)=6x{e}^{x^2}-6x{e}^{x^2}+\frac{1}{3}\left(3\right)-0=1$