Find (d³y)/(dx³) at the point (1,1) for the curve

Find $\dfrac{d^{3}y}{dx^{3}}$ at the point $(1,1)$ for the curve $x^{2}+2xy+y^{2}-4x+2y = 2$.

Step 1: Recognise the perfect square

The first three terms are $\bigl(x+y\bigr)^{2}$, so with $s = x+y$ the equation is

\[ s^{2}-4x+2y = 2, \qquad s' = 1+y'. \tag{1} \]

The point is on the curve: $1+2+1-4+2 = 2$.

Step 2: First derivative

\[ 2s\bigl(1+y'\bigr)-4+2y' = 0 \;\Longrightarrow\; y'\bigl(2s+2\bigr) = 4-2s \;\Longrightarrow\; y' = \frac{2-s}{s+1}. \tag{2} \]

At $(1,1)$, where $s = 2$: $y' = 0$ — a horizontal tangent.

Step 3: Second derivative

Differentiating (2), the two brackets combine into a constant: \[y'' = \frac{-s'\bigl(s+1\bigr)-\bigl(2-s\bigr)s'}{\bigl(s+1\bigr)^{2}} = \frac{-s'\Bigl[\bigl(s+1\bigr)+\bigl(2-s\bigr)\Bigr]}{\bigl(s+1\bigr)^{2}} = \frac{-3\bigl(1+y'\bigr)}{\bigl(s+1\bigr)^{2}}. \tag{3}\]

At $(1,1)$: $y'' = \dfrac{-3\cdot1}{9} = -\dfrac13$.

Step 4: Third derivative

Differentiate (3), again with $s' = 1+y'$: \[y''' = \frac{-3y''\bigl(s+1\bigr)^{2}+6\bigl(1+y'\bigr)^{2}\bigl(s+1\bigr)}{\bigl(s+1\bigr)^{4}} = \frac{-3y''\bigl(s+1\bigr)+6\bigl(1+y'\bigr)^{2}}{\bigl(s+1\bigr)^{3}}. \tag{4}\]

Step 5: Evaluate at $(1,1)$

Here $s+1 = 3$, $y' = 0$, $y'' = -\tfrac13$: \[y''' = \frac{-3\left(-\dfrac13\right)\cdot3+6\cdot1}{27} = \frac{3+6}{27} = \frac13.\]

Step 6: Verification

\[\text{Point on curve: } 1+2+1-4+2 = 2 \;\;\checkmark\]

\[\text{Symbolic implicit differentiation gives } \bigl(0,\,-\tfrac13,\,\tfrac13\bigr) \text{ at } (1,1) \;\;\checkmark\]

\[y' = 0 \text{ and } y''\lt 0 \;\Longrightarrow\; (1,1) \text{ is a local maximum of the branch} \;\;\checkmark\]