Simplify the expression. $\frac{7y-y^{-1}}{3-y^{-2}}$

A) $\frac{y\left(7y^2-1\right)}{3y^2-1},y\ne 0$
B) $\frac{6y^2}{3y^2-1},y\ne 0$
C) $\frac{3y}{7y-1},y\ne 0$
D) $\frac{2y^2}{3y^2-1},y\ne 0$
E) $\frac{7y}{3y-1},y\ne 0$

To simplify this expression, we need to get rid of the negative exponent in the numerator and the denominator. We can do this by multiplying both the numerator and denominator by $y$ to get: $\frac{(7y^2-1)}{(3y-y^3)}$. Then, we can factor out a $y^2-1$ from the numerator and a $1-y^2$ from the denominator to get $\frac{y(7y^2-1)}{(3-y)(1+y)(1-y)}$. Finally, we can simplify by canceling out the $(1-y)$ term in the denominator with the $y-1$ term in the numerator, leaving us with $\frac{y(7y^2-1)}{3-y^2}$, which is equivalent to choice A.