Answer: At \(x=1\), \(y=-1\). Substituting into \(x^{2}+y^{2}+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=29\) gives

\(1^{2}+(-1)^{2}+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=29\)

so \(2+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=29\), hence \(\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=27\).

Therefore \(\frac{\mathrm{d}y}{\mathrm{d}x}=3\).

Differentiate the given equation with respect to \(x\):

\(2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}+3\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\).

Now substitute \(x=1\), \(y=-1\), and \(\frac{\mathrm{d}y}{\mathrm{d}x}=3\):

\(2(1)+2(-1)(3)+3(3)^{2}\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\)

\(-4+27\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\)

so \(27\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=4\), and therefore \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\frac{4}{27}\).

Hence \(\frac{\mathrm{d}y}{\mathrm{d}x}=3\) and \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\frac{4}{27}\) when \(x=1\).

When \(x=1\), we are told that \(y=-1\). Substitute these values into the equation

\(x^{2}+y^{2}+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=29\).

This gives

\(1^{2}+(-1)^{2}+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=29\)

\(2+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=29\)

so

\(\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}=27\).

Therefore

\(\frac{\mathrm{d}y}{\mathrm{d}x}=3\).

Now differentiate the original equation with respect to \(x\):

\(\frac{\mathrm{d}}{\mathrm{d}x}(x^{2})+\frac{\mathrm{d}}{\mathrm{d}x}(y^{2})+\frac{\mathrm{d}}{\mathrm{d}x}\left(\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{3}\right)=0\).

Using the chain rule,

\(2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}+3\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\).

Substitute \(x=1\), \(y=-1\), and \(\frac{\mathrm{d}y}{\mathrm{d}x}=3\):

\(2(1)+2(-1)(3)+3(3)^{2}\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\).

So

\(2-6+27\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\)

\(-4+27\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\).

Hence

\(27\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=4\)

and therefore

\(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\frac{4}{27}\).