Answer
$$ f'(x)= 2(x+1)e^{x^2+2x-3}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(x^n)'=nx^{n-1}$
Since we have
$$ f(x)= e^{x^2+2x-3}$$
then the derivative $ f'(x)$, using the chain rule, is given by
$$ f'(x)=e^{x^2+2x-3}(x^2+2x-3)'=2(x+1)e^{x^2+2x-3}.$$
