f is differentiable at a implies that f'(a) = [f(a+h) - f(a)]/h as h->0
take g(x) = [f(x)]^2
therefore lim h->0 [[f(a+h)]^2 - [f(a)]^2 ]/h = f'(a) * [f(a+h)+f(a)]
and u can easily see that the second bracket is equal to 2*f(a) as h->0
so the square of f is differentiable at a and the derivative is 2f(a)*f'(a)