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

$\text{ we have }f\left(x\right)=3e^{x^2},\text{ differentiating w.r.t.x,we get }$

$\begin{array}{l}{f}^1\left(x\right)=6x{e}^{x^2}\\ f\left(0\right)=3e^0=3,f^1\left(0\right)=6.0\cdot e^0=0\\ f^1\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^1\left(0\right)\\ =6x{e}^{x^2}-6x{e}^{x^2}+\frac{1}{3}\left(3\right)-0=1\end{array}$