Directional derivative of a function Feel like I may have gone wrong somewhere with this question:
Find the directional derivative of the function $f(x,y) = \displaystyle\dfrac{2x}{x-y}$ at the point $P(1, 0)$ in the direction of the vector $v=(4, 3)$.
I got: $f_x(x, y) = \dfrac{-2y}{(x-y)^2}$ $f_y(x, y) = \dfrac{2x}{(x-y)^2}$ 
$D_v(x, y) = 4f_x(x,y)+3f_y(x,y) = \dfrac{6x-8}{(x-y)^2}$ 
$D_v(1, 0) = \dfrac{6-8}{1} = -2$ 
Should I have normalised the vector to $v = (\dfrac{4}{5},\dfrac{3}{5})$ so the answer would be $\dfrac{-2}{5}$?

You made a mistake in the second-to-last step. You should get $-8y+6x$ instead of $-8+6x$ in the numerator.
Other than that your approach and calculations look fine to me.
