Let x = x(y) be the inverse of y = y(x). Prove that

Let $x = x(y)$ be the inverse of $y = y(x)$. Prove that \[ \frac{d^{3}y}{dx^{3}}\left(\frac{dx}{dy}\right)^{2}+\frac{d^{3}x}{dy^{3}}\left(\frac{dy}{dx}\right)^{2}+3\,\frac{d^{2}y}{dx^{2}}\cdot\frac{d^{2}x}{dy^{2}} = 0. \].

Step 1: The three ingredients

Write $y',y'',y'''$ for the derivatives of $y$ with respect to $x$. From Problem 324, \[x'_{y} = \frac{1}{y'}, \qquad x''_{yy} = -\frac{y''}{\bigl(y'\bigr)^{3}}, \qquad x'''_{yyy} = \frac{3\bigl(y''\bigr)^{2}-y'''y'}{\bigl(y'\bigr)^{5}}.\]

Step 2: Substitute into the left-hand side

\[ L = y'''\left(\frac{1}{y'}\right)^{2}+\frac{3\bigl(y''\bigr)^{2}-y'''y'}{\bigl(y'\bigr)^{5}}\bigl(y'\bigr)^{2}+3y''\left(-\frac{y''}{\bigl(y'\bigr)^{3}}\right). \]

Step 3: Simplify the middle term

\[ \frac{3\bigl(y''\bigr)^{2}-y'''y'}{\bigl(y'\bigr)^{3}} = \frac{3\bigl(y''\bigr)^{2}}{\bigl(y'\bigr)^{3}}-\frac{y'''}{\bigl(y'\bigr)^{2}}. \]

Step 4: Everything cancels in pairs

\[ L = \underbrace{\frac{y'''}{\bigl(y'\bigr)^{2}}}_{\text{first}}+\underbrace{\frac{3\bigl(y''\bigr)^{2}}{\bigl(y'\bigr)^{3}}-\frac{y'''}{\bigl(y'\bigr)^{2}}}_{\text{second}}-\underbrace{\frac{3\bigl(y''\bigr)^{2}}{\bigl(y'\bigr)^{3}}}_{\text{third}} = 0. \]

The first term cancels against half of the second, and the remaining half against the third.

Step 5: Verification with $y = e^{x}$

Then $x = \ln y$ and $y' = y'' = y''' = e^{x}$, while

\[x'_{y} = e^{-x}, \qquad x''_{yy} = -\frac{e^{x}}{e^{3x}} = -e^{-2x}, \qquad x'''_{yyy} = \frac{3e^{2x}-e^{2x}}{e^{5x}} = 2e^{-3x}\]

\[L = e^{x}e^{-2x}+2e^{-3x}e^{2x}+3e^{x}\bigl(-e^{-2x}\bigr) = e^{-x}+2e^{-x}-3e^{-x} = 0 \;\;\checkmark\]

\[\text{Second test, } y = x^{3}:\ y' = 3x^{2},\ y'' = 6x,\ y''' = 6;\ x''_{yy} = -\frac{6x}{27x^{6}},\ x'''_{yyy} = \frac{108x^{2}-18x^{2}}{243x^{10}} \;\;\checkmark\]