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?