Answer: \(\frac{\mathrm dy}{\mathrm dx}=-\frac34,\qquad \frac{\mathrm d^2y}{\mathrm dx^2}=\frac{9}{32}\quad\text{at }(1,0).\)

The curve is

\(xy+(x+y)^3=1.\)

Differentiate implicitly:

\(y+x\frac{\mathrm dy}{\mathrm dx}+3(x+y)^2\left(1+\frac{\mathrm dy}{\mathrm dx}\right)=0.\)

At \((1,0)\), this becomes

\(0+\frac{\mathrm dy}{\mathrm dx}+3(1)^2\left(1+\frac{\mathrm dy}{\mathrm dx}\right)=0.\)

Hence

\(4\frac{\mathrm dy}{\mathrm dx}+3=0,\qquad \frac{\mathrm dy}{\mathrm dx}=-\frac34.\)

Differentiate

\(y+xy'+3(x+y)^2(1+y')=0\)

again with respect to \(x\). This gives

\(2y'+xy''+6(x+y)(1+y')^2+3(x+y)^2y''=0.\)

At \((1,0)\), with \(y'=-\frac34\),

\(2\left(-\frac34\right)+y''+6(1)\left(1-\frac34\right)^2+3y''=0.\)

So

\(-\frac32+6\cdot\frac1{16}+4y''=0,\)

\(-\frac32+\frac38+4y''=0.\)

Thus

\(4y''=\frac98,\qquad y''=\frac{9}{32}.\)