For fixed y, let \(\displaystyle g(x) = f(x+y)-f(x)-f(y)\). Then \(\displaystyle g(0) = 0\). Show that \(\displaystyle g'(x)>0\) for all x>0, and deduce that \(\displaystyle g(x)>0\) for all x>0.

The only reason that you need the condition f''(x)>0 (for x>0) is that it implies that f'(x) is a strictly increasing function, so that f'(x+y) > f'(x) when y>0.

The value of f''(x) when x=0 is not relevant. The fact that f''(x) > 0 when x>0 does in fact imply that f is strictly convex.