Simplifying the second derivative

Question

I need to simplify the below with respect to $y''$. I'm given: $$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}$$

The final result should look like: $$-\frac{3x^2(y^4+x^4)}{y^7}$$

Here are my steps:

1. Combine term: $-3x^3y^2\left(-\frac{x^3}{y^3}\right)$. $$-3x^3y^2\left(-\frac{x^3}{y^3}\right)=\frac{3x^6y^2}{y^3}$$ Our expression is now: $$y''=-\frac{3x^2y^3+\frac{3x^6y^2}{y^3}}{y^6}$$
2. Multiply the final term in the numerator, $\frac{3x^6y^2}{y^3}$, by the it's reciprocal in the denominator, $\frac{y^6}{1}$, to eliminate the complex fraction. $$\frac{3x^6y^2}{y^3} * \frac{y^6}{1}=\frac{3x^6y^8}{y^3}$$ Our expression is now: $$y''=-\frac{3x^2y^3}{y^6}+\frac{3x^6y^8}{y^3}$$
3. Multiply term $\frac{3x^6y^8}{y^3}$ by $\frac{y^3}{y^3}$ to make common denominators. $$\frac{3x^6y^8}{y^3}*\frac{y^3}{y^3}=\frac{3x^6y^{11}}{y^6}$$ Our separate terms can now be added, and our expression will look like: $$y''=-\frac{3x^2y^3+3x^6y^{11}}{y^6}$$
4. Now factor the numerator, our expression will become: $$y''=-\frac{3x^2y^3(1+x^4y^{8})}{y^6}$$

Have I made a mistake? I'm not sure where to go from here. I need to achieve this expression as stated at the top my question: $$y''=-\frac{3x^2(y^4+x^4)}{y^7}$$

Answer 1

$$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}=-\frac{3x^2y^3+3x^3\left(\frac{x^3}{y}\right)}{y^6}=y''=-\frac{\left(\frac{3x^2y^4+3x^6}{y}\right)}{y^6}=-\frac{3x^2y^4+3x^6}{y^7}=-\frac{3x^2(y^4+x^4)}{y^7}$$