What is $\frac{dy}{dx}|_{y=-1}$ for $(xy^3 + x^2y^7)\frac{dy}{dx} = 1$ given that $y \left(\frac{1}{4}\right)=1$ 
Suppose a solution of the differential equation
  $$(xy^3 + x^2y^7)\frac{dy}{dx} = 1$$
  satisfies the initial condition $y \left(\frac{1}{4}\right)=1$ . Then the value of $\dfrac{dy}{dx}$ when $y = −1$ is
(A) $4/3$
(B) $−4/3 $
(C) $16/5$ 
(D) $−16/5$.    

This is not a homogeneus equation so I can't solve it. Do I need to know some special procedure to solve this problem? If the main differential equation is solved then I can solve the problem.
 A: Since you want to know $x$ at given values of $y$, it's more convenient to switch dependent and independent variables and write the problem as
$$ \dfrac{dx}{dy} = xy^3+x^2 y^7, \ x(1) = 1/4 $$
Now this Bernoulli differential equation actually does have a closed-form solution
$$ x \left( y \right) = \left( {\rm e}^{(1-y^4)/4}+4-{y}^{4}
 \right) ^{-1}
$$
But even without that, the fact that the right side of the differential equation
is an odd function of $y$ shows that all solutions that pass through $y=0$
must be even functions of $y$.  Thus we must have $x(-1) = x(1) = 1/4$.
And then you can calculate $dy/dx$ at the point $(x=1/4, y=-1)$ from the original
differential equation.
In terms of the original differential equation, of course, we can't have $y$ as a function of $x$ with both $y(1/4) = 1$ and $y(1/4) = -1$.  What it means is that the two solutions with initial conditions $y(1/4) = 1$ and $y(1/4) = -1$ collide at a singularity on the line $y=0$, but nevertheless we can regard them as two branches of the same integral curve.  

A: This is, I think, a neat little problem, so I will provide a more in-depth explanation than my comment. It still smells like homework, so I won't solve it for you, just in case.
The problem states that $y(1/4) = 1$, meaning that when $x = 1/4$, $y = 1$. In that case, we can plug in those values to obtain $\frac{dy}{dx}\mid_{x=1/4} = 16/5$. But that doesn't solve our problem, yet.
Next, we want to figure out what happens when $y=-1$. We don't know what the corresponding value of $x$ is, but we can still plug in $y=-1$ to obtain $\left(x(-1)^3+x^2(-1)^7\right)\frac{dy}{dx}\mid_{y=-1} = 1$.
Now, it's multiple choice, which makes our life easier. We want to turn it into a quadratic equation.
First, multiply both sides by $1/\frac{dy}{dx}\mid_{y=-1}$, and then bring over the $x$'s:
$0 = x^2+x+\frac{1}{\frac{dy}{dx}\mid_{y=-1}}$.
Let $c = \frac{1}{\frac{dy}{dx}\mid_{y=-1}}$. Now, let's use the quadratic equation:
$x = -\frac{1}{2}\pm \frac{\sqrt{1-4c}}{2}$.
Plug in your multiple choice values for $c$.
Immediately, you should be able to eliminate two of them. Why?
For the next two, you should arrive at a contradiction with one of them. Can you see what it is, and why?
A: Maybe not too useful, but the solution for y in terms of x is expressible in terms of the Lambert W function, defined implicitly by $$x=W(x) e^{W(x)}$$ 
The general solution is expressible as either
$$ y= \left[ -4 \ln Q - 4 \ln\left(\frac{4}{A}\right) \right] ^{1/4} $$
Where $$Q = W \frac{A}{4} e^{{1/4x} -1}$$ 
And $A$ is an arbitrary constant.
Or as 
$$ y = \left[ 4- \frac{1}{x} + 4 Q \right] ^{1/4} $$
The latter is the form given by Wolfram alpha, which is why I think it worth adding to the previous discussions 
