The constant term 3 cancels out. What remains is the difference of two squares: 4^2-(4+h)^2=(4-4-h)(4+4+h)=-h(8+h). Divide by h: -(8+h). h is very small compared to 8 so the limit is -8.
f(x)=3-x^2; f'(x)=-2x. If -2x=-8, x=4, so the point is (4,3-16)=(3,-13).