Tricky Differential Equation Problem I am unsure of how to tackle the following differential equation:

$$ dx+ x\,dy = e^{-y}\sec^2y\,dy$$

I have done the following so far:

$$dx + x\,dy = e^{-y} \sec^2y \, dy$$

$$=&gt;dx = e^{-y} \sec^2y \, dy - x \, dy$$

$$=&gt;dx = (e^{-y} \sec^2y - x) \, dy$$

$$=&gt;dx/dy = (e^{-y}) \sec^2y - x $$

Is this the correct approach? How can the problem be solved after this?

(Not homework: preparing for a test)
Hint: Rewrite/Simplify it as:

$$e^y~dx + (x e^y - \sec^2 y)~ dy = 0$$

Now, check this as being an Exact Equation to proceed.
Yeah! But if you just want to, you can also use the other method which is Linear Differential equation, where in the standard solution goes like this: dx/dy+xP(y)=Q(y)--> if eq. is linear in x; dy/dx+yP(x)=Q(x)--> if eq. is linear in y. 

since the given problem is linear in x, use dx/dy+xP(y)=Q(y)
where P(y)=1 and Q(y)=e^(−y)sec^(2)y
and IF (Integrating factor): e^(integral of P(y) or e enter image description here

then use the formula:
x*IF=enter image description here(IF * Q(y))dy + C
thus,
x*e^(y)=tan^(2)y+C [this is the final answer in general form ]