Differentiating: 2x-xdy/dx-y-2ydy/dx=0. At (2,5), 4-2dy/dx-5-10dy/dx=0, 12dy/dx=-1, dy/dx=-1/12.
However, (2,5) does not lie on the curve so the gradient has no meaning.
When x=2 and y=5, x2-xy-y2=4-10-25=-31, so the gradient only has meaning if x2-xy-y2=-31.
However, if the question means the gradient dy/dx when a tangent passes through (2,5) we can write:
dy/dx=(2x-y)/(x+2y). At point (p,q) on the curve, dy/dx=(2p-q)/(p+2q) and this is the gradient m of the line y=mx+a passing through (2,5) and (p,q).
y-5=m(x-2), y=mx-2m+5 so a=-2m+5, or m=(y-5)/(x-2).
Since (p,q) lies on the curve, p2-pq-q2-15=0, p=(q±√(5q2+60))/2, using the quadratic formula.
2p-q=±√(5q2+60), p+2q=(5q±√(5q2+60))/2. We then need to substitute for p in (2p-q)/(p+2q) so the gradient is in one variable, q. We have m=(q-5)/(p-2) from the equation of the tangent line since (p,q) must lie on the line as well as the curve. By equating these two equations for m and substituting for p we get:
(2p-q)/(p+2q)=(q-5)/(p-2), ±2√(5q2+60)/(5q±√(5q2+60))=2(q-5)/(q-4±√(5q2+60)).
The algebra gets too complicated after this. However, q=-2.04041 and -2.79830 approx, so p=-5.5151 and 3.5796, making (p,q)=(-5.5151,-2.0404) and (3.5796,-2.7983).
dy/dx=m=7.0404/7.5151=0.9368 and -7.7983/1.5796=-4.9369.
Because of this complication, it is unlikely that this is the intended solution, and perhaps the question has been wrongly presented.