A ground to ground projectile is at a point A at $t=\frac T3$ and at B at $t=\frac{5T}{6}$ and reaches the ground at $t=T$. Find the difference in height between A and B.
Time of flight is $$T=\frac{2u_y}{g}$$ where $u_y$ is vertical component of initial velocity
Therefore
$$t_a=\frac{2u_y}{3g}$$and $$t_b=\frac{5u_y}{3g}$$
Then $$h_a=u_y t_a-\frac 12 g t_a^2$$ $$h_b=u_yt_b-\frac 12 g t_b^2$$
$$h_b-h_a=u_y(t_b-t_a)-\frac 12 g (t_b^2-t_a^2)$$ $$=(t_b-t_a)(u_y-\frac 12 g (t_b+t_a))$$ $$=\frac {u_y}{g}(u_y-\frac 12 g(\frac{7u_y}{3g}))$$
So I am getting a negative value, here. What am I doing wrong?
You are doing nothing wrong! Its totally fine that $h_b - h_a<0$. This simply means that $h_b<h_a$. That is, that in $A$ the projectile is higher than in $B$, which is intuitive, since in point $B$, the projectile is near from hitting the ground.