If you are going to use the Product Rule, you have \begin{align*} H'(u) &= \left(u-\sqrt{u}\right)'\left(u-\sqrt{u}\right) + \left(u-\sqrt{u}\right)\left(u-\sqrt{u}\right)'\\ &= \left(u - u^{1/2}\right)'\left(u-u^{1/2}\right) + \left(u-u^{1/2}\right)\left(u-u^{1/2}\right)'\\ &= \left( 1 - \frac{1}{2}u^{-1/2}\right)\left(u - u^{1/2}\right) + \left(u-u^{1/2}\right)\left(1 - \frac{1}{2}u^{-1/2}\right)\\ &= 2\left(u - u^{1/2}\right)\left(1 - \frac{1}{2}u^{-1/2}\right). \end{align*}
The first step is the Product Rule. The second step is just the fact that $\sqrt{u}=u^{1/2}$. The third step uses the Sum Rule and the Power Rule. The fourth and final step is just the fact that the two summands are equal.
If you want to further multiply out the product, we have: \begin{align*} H'(u) &= 2\left(u - u^{1/2}\right)\left(1 - \frac{1}{2}u^{-1/2}\right)\\ &= \left(u - u^{1/2}\right)\left(2 - u^{-1/2}\right)\\ &= 2u - uu^{-1/2} - 2u^{1/2} + u^{1/2}u^{-1/2}\\ &= 2u - u^{1/2} -2u^{1/2}+1\\ &= 2u-3u^{1/2} + 1. \end{align*}
For the function you identify in the comments as the "correct one": $H(u) = (u-\sqrt{u})(u+\sqrt{u})$ assuming you want to exercise the Product Rule, we have: \begin{align*} H'(u_) &= \left( u - \sqrt{u}\right)'\left(u+\sqrt{u}\right) + \left(u-\sqrt{u}\right)\left(u+\sqrt{u}\right)'\\ &= \left( u - u^{1/2}\right)'\left(u+u^{1/2}\right) + \left(u-u^{1/2}\right)\left(u+u^{1/2}\right)'\\ &=\left(1 - \frac{1}{2}u^{-1/2}\right)\left(u+u^{1/2}\right) + \left(u - u^{1/2}\right)\left(1 + \frac{1}{2}u^{-1/2}\right)\\ &= u+u^{1/2}-\frac{1}{2}u^{-1/2}u - \frac{1}{2}u^{-1/2}u^{1/2} + u + \frac{1}{2}uu^{-1/2} - u^{1/2} - \frac{1}{2}u^{1/2}u^{-1/2}\\ &= u + u^{1/2} - \frac{1}{2}u^{1/2} - \frac{1}{2} + u + \frac{1}{2}u^{1/2} - u^{1/2}-\frac{1}{2}\\ &= 2u-1. \end{align*} You can verify this is correct, since $(u-\sqrt{u})(u+\sqrt{u}) = u^2 - u,$ and (u^2-u)' = 2u-1.