Answer
\[y''=\frac{2xy^3-2x^4}{y^5}\]
Work Step by Step
\[x^3-y^3=7\]
Differentiate implicitly with respect to $x$
\[3x^2-3y^2(y)'=0\]
\[\Rightarrow y'=\frac{3x^2}{3y^2}\]
\[\Rightarrow y'=\frac{x^2}{y^2}\;\;\;...(1)\]
Differentiating (1) implicitly with respect to $x$
\[\Rightarrow y''=\frac{(2x)y^2-x^2(2y)y'}{y^4}\]
Using (1)
\[\Rightarrow y''=\frac{(2x)y^2-x^2(2y)\left(\frac{x^2}{y^2}\right)}{y^4}\]
\[\Rightarrow y''=\frac{2xy^3-2x^4}{y^5}\]
Hence , \[\Rightarrow y''=\frac{2xy^3-2x^4}{y^5}.\]
