Find a unit vector in the direction in which f decreases most rapidly at P Find a unit vector in the direction in which f decreases most rapidly at P, and find the rate of change of f at P in that direction:
$f(x, y) = \sqrt{\frac{x-y}{x+y}} $ ; P(3,1)
I worked out $\nabla f(x,y) = (\frac{\sqrt{2}}{16}, -\frac{3\sqrt{2}}{16} )$ at P(3,1)
And then to find the unit vector in the direction in which f decreases:
$-u= -\frac{8}{15}(\frac{\sqrt{2}}{16}, -\frac{3\sqrt{2}}{16} ) = (-\frac{\sqrt{10}}{10},-3\frac{\sqrt{10}}{10})$ 
$D_uf(3,1) = \| \nabla f(3,1) \| = \frac{\sqrt{5}}{8}$
A tutor advised me that this is incorrect however I'm not sure where I'm going wrong. 

The unit vector with direction the gradient is: 
$$u={\nabla f\over||\nabla f||}.$$
So you want $-u$. Comparing two values in your work ${8\over15}$ and ${\sqrt5\over8}$ makes me skeptical. 
For the rate of change in direction of $-u$, we write
$$D_{-u}f=\nabla f\cdot(-u)=-\nabla f\cdot{\nabla f\over||\nabla f||}=-{||\nabla f||^2\over||\nabla f||}=-||\nabla f||.$$
