if a curve y=f(x) passes through the point (1,-1)satisfies the differential equation y(1+xy) dx= x dy then f(-1/2)

write it as

ydx + xy^2dx= xdy

or (ydx – xdy)/y^2 = – xdx

now note that (ydx – xdy)/y^2 = d(x/y)

so d(x/y)= – xdx

integrate both sides

x/y= – x^2/2 + C

Dear student 
y/x (1+xy) = dy/dx 
y = vx = y/x = v 
dy/dx = v+xdv/dx 
v(1+vx^2) = v+xdv/dx 
v^2x^2 = xdv/dx 
Integrate this 
x^2/2 = -1/v + c 
x^2/2 = -x/y - 1/2 
f(-1/2) , put x = -1/2 
1/8 = 1/2y - 1/2 
y = 4/5