<s>[INST] Simplifying the second derivative 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:


*

*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:
$$y''=-\frac{3x^2y^3+\frac{3x^6y^2}{y^3}}{y^6}$$

*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:
$$y''=-\frac{3x^2y^3}{y^6}+\frac{3x^6y^8}{y^3}$$

*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:
$$y''=-\frac{3x^2y^3+3x^6y^{11}}{y^6}$$

*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:
$$y''=-\frac{3x^2(y^4+x^4)}{y^7}$$
 [/INST] $$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}=-\frac{3x^2y^3+3x^3\left(\frac{x^3}{y}\right)}{y^6}=y''=-\frac{\left(\frac{3x^2y^4+3x^6}{y}\right)}{y^6}=-\frac{3x^2y^4+3x^6}{y^7}=-\frac{3x^2(y^4+x^4)}{y^7}$$
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