From part (a), we have
2 x + 3 y 2 d y d x − 2 y − 2 x d y d x = 0 {\displaystyle {\begin{aligned}2x+3y^{2}{\frac {dy}{dx}}-2y-2x{\frac {dy}{dx}}=0\end{aligned}}}
Differentiating again gives
2 + 3 ( 2 y ( d y d x ) 2 + y 2 d 2 y d x 2 ) − 2 d y d x − 2 ( d y d x + x d 2 y d x 2 ) = 0 {\displaystyle {\begin{aligned}2+3\left(2y\left({\frac {dy}{dx}}\right)^{2}+y^{2}{\frac {d^{2}y}{dx^{2}}}\right)-2{\frac {dy}{dx}}-2\left({\frac {dy}{dx}}+x{\frac {d^{2}y}{dx^{2}}}\right)=0\end{aligned}}}
Isolating for the second derivative gives
2 + 6 y ( d y d x ) 2 + 3 y 2 d 2 y d x 2 − 4 d y d x − 2 x d 2 y d x 2 = 0 2 − 4 d y d x + 6 y ( d y d x ) 2 + ( 3 y 2 − 2 x ) d 2 y d x 2 = 0 d 2 y d x 2 = − 2 + 4 d y d x − 6 y ( d y d x ) 2 3 y 2 − 2 x {\displaystyle {\begin{aligned}2+6y\left({\frac {dy}{dx}}\right)^{2}+3y^{2}{\frac {d^{2}y}{dx^{2}}}-4{\frac {dy}{dx}}-2x{\frac {d^{2}y}{dx^{2}}}&=0\\2-4{\frac {dy}{dx}}+6y\left({\frac {dy}{dx}}\right)^{2}+(3y^{2}-2x){\frac {d^{2}y}{dx^{2}}}&=0\\{\frac {d^{2}y}{dx^{2}}}&={\frac {-2+4{\frac {dy}{dx}}-6y\left({\frac {dy}{dx}}\right)^{2}}{3y^{2}-2x}}\end{aligned}}}
Now we plug in x = y = 1 {\displaystyle x=y=1} into this equation. From part (a), we know that the derivative at 1 is 0 and thus, we get
d 2 y d x 2 = − 2 + 4 ( 0 ) − 6 ( 1 ) ( 0 ) 2 3 ( 1 ) 2 − 2 ( 1 ) = − 2 1 = − 2 {\displaystyle {\begin{aligned}{\frac {d^{2}y}{dx^{2}}}&={\frac {-2+4(0)-6(1)(0)^{2}}{3(1)^{2}-2(1)}}={\frac {-2}{1}}=-2\end{aligned}}}
and so f ″ ( 1 ) = − 2 {\displaystyle \displaystyle f''(1)=-2}
into this equation. From part (a), we know that the derivative at 1 is 0 and thus, we get
