myT=x^2-2 x y;
myCP=ContourPlot[myT, {x,0,1},{y,0,1}, Contours->40];
myPoint= Graphics[{PointSize[Large],Black, Point[{1/2,1/3}]}];
myBound=Plot[x,{x,0,1}];
Thickness[0.05];
myQuad=ParametricPlot[{t,t^2},{t,0,1}];
Show[myCP, myPoint, myBound, myQuad]

c.)To estimate the coordinates (in Mathematica), right click the picture and choose “Get Coordinates”.

d.) 
“the cat moves in such a way that for each change in its x direction of 2 units, the change in the y direction is –1 unit”.

Call (OverVector[u])=⟨2,-1⟩. We need a unit vector pointing in this direction:

(OverVector[u])={2,-1};
(OverHat[u])=(OverVector[u])/Norm[(OverVector[u])]

Now we’re ready to use the expression for the directional derivative: (Subscript[D, (OverHat[u])])(OverVector[∇])T·(OverHat[u]) if we first calculate the gradient:

gradT={   D[myT,x], D[myT, y]   }

gradAtCat=gradT /.{x->1/2,   y->1/3}

{gradAtCat. (OverHat[u]), N[gradAtCat. (OverHat[u])]}

{((Sqrt[5])/3),0.7453559924999299`}

Visualizing these vectors...

myUhat= Graphics[{Blue,  Arrow[
{{1/2,1/3},{1/2,1/3}+(OverHat[u])}
]}];
mygradT=Graphics[{Red,  Arrow[
{{1/2,1/3},{1/2,1/3}+gradAtCat}
]}];
Show[myCP, myPoint, myBound, myQuad,myUhat,mygradT]

Try this to find the exact coordinates of the maximum point along the border:

We already have parametric equations for the bottom boundary: x(t)=t; y(t)=(t^2).

Substitute these into the expression for temperature T(x,y)=(x^2)-2x y, to find T(t).

Find the value of t which maximizes T(t).

Substitute that back into expression x(t), and y(t) to find the coordinates of the maximum.

Plot[  t^2-2t^3, {t,0,1}]

Plot[  t^2-2t^3, {t,0,1}, PlotRange->{-.1,.1}]