# What is the equation of the normal line of f(x)=(x-3e^x)^(1/2)-x^2 at x = -3?

A normal can't be found as $f \left(x\right)$ doesn't exist as the square root in the formula is always negative and you can't have a negative square root (that is real at least). graph{y=x-3e^x [-40.73, 38.1, -26.78, 12.62]}
The domain of $y = \sqrt{x}$ is $x > 0$, but the range of $y = x - 3 {e}^{x}$ is $y \le - 2.099$ (3 d.p) - it is always negative.