If a curve y=f(x) passes through the point (1,1) and satisfies the differential equation: y(1+xy)dx=xdy, then f(2) is equal to:

y=f(x) passes through (1,1)
y(1+xy)dx=xdy
ydx+xy^2dx=xdy
ydx-xdy=-xy^2dx
(ydx-xdy)/y^2=-xdx
∫(ydx-xdy)/y^2=∫(-x)dx
1/y=-x^2/2+C
passes through point (1,1)
1/1=-1/2+C
C=3/2
1/y=-x^2/2+3/2
y=2/(3-x^2)
y=f(x)
f(2)=2/(3-2^2)=2/(3-4)=-2
However, there appears to be an error in either the problem statement or the provided solution. Let's re-examine the solution:
ydx - xdy = -xy^2dx
d(x/y) = -xdx
Integrating both sides:
∫d(x/y) = ∫-xdx
x/y = -x^2/2 + C
Since the curve passes through (1, 1):
1/1 = -1/2 + C
C = 3/2
x/y = -x^2/2 + 3/2
y = x/(-x^2/2 + 3/2) = 2x/(3-x^2)
f(2) = 2(2)/(3-2^2) = 4/(3-4) = -4
Let's check the original differential equation:
y(1+xy)dx = xdy
ydx + xy^2dx = xdy
ydx - xdy = -xy^2dx
Dividing by y^2:
(ydx - xdy)/y^2 = -xdx
d(x/y) = -xdx
Integrating both sides:
∫d(x/y) = ∫-xdx
x/y = -x^2/2 + C
Using the point (1,1):
1 = -1/2 + C
C = 3/2
x/y = -x^2/2 + 3/2
y = 2x/(3-x^2)
f(2) = 2(2)/(3-4) = -4
There might be a mistake in the options or the provided solution. The correct answer based on the given differential equation and point is f(2) = -4. This value isn't among the options.