Simplifying the second derivative

I need to simplify the below with respect to ${y}^{\u2033}$. I'm given:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}-3{x}^{3}{y}^{2}(-\frac{{x}^{3}}{{y}^{3}})}{{y}^{6}}$

The final result should look like:

$-\frac{3{x}^{2}({y}^{4}+{x}^{4})}{{y}^{7}}$

Here are my steps:

1.Combine term: $-3{x}^{3}{y}^{2}(-\frac{{x}^{3}}{{y}^{3}})$

$-3{x}^{3}{y}^{2}(-\frac{{x}^{3}}{{y}^{3}})=\frac{3{x}^{6}{y}^{2}}{{y}^{3}}$

Our expression is now:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}+\frac{3{x}^{6}{y}^{2}}{{y}^{3}}}{{y}^{6}}$

2.Multiply the final term in the numerator, $\frac{3{x}^{6}{y}^{2}}{{y}^{3}}$, by the it's reciprocal in the denominator, $\frac{{y}^{6}}{1}$, to eliminate the complex fraction.

$\frac{3{x}^{6}{y}^{2}}{{y}^{3}}\ast \frac{{y}^{6}}{1}=\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$

Our expression is now:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}}{{y}^{6}}+\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$

3.Multiply term $\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$ by $\frac{{y}^{3}}{{y}^{3}}$ to make common denominators.

$\frac{3{x}^{6}{y}^{8}}{{y}^{3}}\ast \frac{{y}^{3}}{{y}^{3}}=\frac{3{x}^{6}{y}^{11}}{{y}^{6}}$

Our separate terms can now be added, and our expression will look like:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}+3{x}^{6}{y}^{11}}{{y}^{6}}$

4.Now factor the numerator, our expression will become:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}(1+{x}^{4}{y}^{8})}{{y}^{6}}$

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:

${y}^{\u2033}=-\frac{3{x}^{2}({y}^{4}+{x}^{4})}{{y}^{7}}$

I need to simplify the below with respect to ${y}^{\u2033}$. I'm given:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}-3{x}^{3}{y}^{2}(-\frac{{x}^{3}}{{y}^{3}})}{{y}^{6}}$

The final result should look like:

$-\frac{3{x}^{2}({y}^{4}+{x}^{4})}{{y}^{7}}$

Here are my steps:

1.Combine term: $-3{x}^{3}{y}^{2}(-\frac{{x}^{3}}{{y}^{3}})$

$-3{x}^{3}{y}^{2}(-\frac{{x}^{3}}{{y}^{3}})=\frac{3{x}^{6}{y}^{2}}{{y}^{3}}$

Our expression is now:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}+\frac{3{x}^{6}{y}^{2}}{{y}^{3}}}{{y}^{6}}$

2.Multiply the final term in the numerator, $\frac{3{x}^{6}{y}^{2}}{{y}^{3}}$, by the it's reciprocal in the denominator, $\frac{{y}^{6}}{1}$, to eliminate the complex fraction.

$\frac{3{x}^{6}{y}^{2}}{{y}^{3}}\ast \frac{{y}^{6}}{1}=\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$

Our expression is now:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}}{{y}^{6}}+\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$

3.Multiply term $\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$ by $\frac{{y}^{3}}{{y}^{3}}$ to make common denominators.

$\frac{3{x}^{6}{y}^{8}}{{y}^{3}}\ast \frac{{y}^{3}}{{y}^{3}}=\frac{3{x}^{6}{y}^{11}}{{y}^{6}}$

Our separate terms can now be added, and our expression will look like:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}+3{x}^{6}{y}^{11}}{{y}^{6}}$

4.Now factor the numerator, our expression will become:

${y}^{\u2033}=-\frac{3{x}^{2}{y}^{3}(1+{x}^{4}{y}^{8})}{{y}^{6}}$

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:

${y}^{\u2033}=-\frac{3{x}^{2}({y}^{4}+{x}^{4})}{{y}^{7}}$