Find the value of $x$ for which the gradient is zero. Given that $y=(x-3) \sqrt{x-1} $. Find the value of $x$ for which the gradient is zero.
The derivative of the function is $$\frac{dy}{dx} =\frac{x-3}{2\sqrt{x-1}} + \sqrt{x-1}.$$

Your derivative $\dfrac{dy}{dx}$ is correct. 
$$\frac{dy}{dx} =\frac{x-3}{2\sqrt{x-1}} + \sqrt{x-1} = \frac{(x-3) +\cdot 2\sqrt{x -1}\cdot \sqrt{x-1}) + }{2\sqrt{x-1}} = \frac{(x-3)+ 2(x-1)}{2\sqrt{x-1}}$$
So, $$\frac{dy}{dx} = 0 \implies (x-3)+ 2(x-1) = 0$$ 
Now simplify: $$(x-3)+2(x-1) = 0 \iff x-3+2x - 2 = 0\iff 3x-5= 0 \iff x = \frac{5}3$$
