Use $ \ \nabla f(3,2) \ $ to find a vector normal to the curve at $ \ (3,2)\ $ View the curve $ \ (y-x)^2+2=xy-3 \ $ as a contour of $ \ f(x,y) \ $
Use $ \ \nabla f(3,2) \ $ to find a vector normal to the curve at $ \ (3,2)\ $
Answer: 
Let $ \ f(x,y)=(y-x)^2-xy+5=0 \ $
Then, 
$ \nabla f(x,y)=\left\langle f_x,f_y \right\rangle \ = \left\langle 2(x-y)-y,2(y-x)-x \right\rangle $
Therefore,
$ \nabla f(3,2)=\left\langle0,-5 \right\rangle \ $ , which is normal t the level curves but not on the curve  $ \ f(x,y) \ $
How to find find the vector normal to the curve $ \ f(x,y) \ $ using $ \ \nabla f(3,2) \ $ ?
Help me doing this.

Here is a graph of your curve, plotted by Maple.
It shows clearly that the curve is horizontal at $(3,2)$, so the normal is vertical, so your answer is correct and the software marking it is wrong.
The only suggestion I could make is that any vertical vector is normal to the curve at this point, that is, any vector $(0,b)$ with $b\ne0$.  Try a unit vector, $(0,1)$ or $(0,-1)$ and see if that gets marked correct.
