Answer: At \((0,1)\),

\(\frac{\mathrm dy}{\mathrm dx}=-\frac13,\qquad \frac{\mathrm d^2y}{\mathrm dx^2}=-\frac49.\)

The curve is

\((x+y)(x^2+y^2)=1.\)

Differentiate implicitly:

\((1+y')(x^2+y^2)+(x+y)(2x+2yy')=0.\)

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

\((1+y')(1)+(1)(2y')=0.\)

So

\(1+3y'=0,\qquad y'=-\frac13.\)

Differentiate the first derivative equation again:

\((x^2+y^2)y''+2(1+y')(2x+2yy')+(x+y)(2+2(y')^2+2yy'')=0.\)

At \(x=0\), \(y=1\), \(y'=-\frac13\), this becomes

\(y''+2\left(1-\frac13\right)\left(0+2\cdot1\cdot-\frac13\right)+1\left(2+2\cdot\frac19+2y''\right)=0.\)

Thus

\(y''+\frac43\left(-\frac23\right)+2+\frac29+2y''=0.\)

So

\(3y''-\frac89+\frac{20}{9}=0,\)

\(3y''+\frac{12}{9}=0,\qquad 3y''+\frac43=0.\)

Hence

\(y''=-\frac49.\)