You can bring this differencial equation to an exact differencial equation by an integrating factor.

Write,

(1/y-e^x)+(1-y^2)y'=0

And let,

M(x,y)=1/y-e^x

N(x,y)=1-y^2

Note the cross partial test fails,

M_y not = N_x

But,

(M_y-N_x)/N

Is a function only of "y" thus, you can find an integrating factor.