Given the function $xe^y+ye^x=0$, we know from implicit function theorem that $\exists C^1$ function $\phi$ s.t. $y=\phi(x)$ is a solution $\forall x$ close to $0$.
How can I show that $\phi$ is $C^2$ on an open interval centered at the origin and find $\phi'' (0)$?
Thank you all.
Using implicit differentiation and noting that $y=y(x)$ due to the implicit function theorem: \begin{align*} \frac{d}{dx} \left[ x e^y + y e^x \right] &= \frac{d}{dx} [0] \\ e^y + x e^y \frac{dy}{dx}+y e^x + \frac{dy}{dx} e^x &= 0. \end{align*} At this stage, we could solve for $\frac{dy}{dx}$ and differentiate using the quotient rule, or we can use implicit differentiation again. \begin{align*} \frac{d}{dx} \left[ e^y + x e^y \frac{dy}{dx}+y e^x + \frac{dy}{dx} e^x\right] = \frac{d}{dx} [0] \\ e^y \frac{dy}{dx} + e^y \frac{dy}{dx} + x e^y \left(\frac{dy}{dx} \right)^2 + x e^y \frac{d^2 y}{dx^2} + \frac{dy}{dx} e^x + y e^x + \frac{d^2y }{dx^2} e^x + \frac{dy}{dx} e^x = 0\\ \frac{d^2y}{dx^2} = \frac{-y e^x - 2 \frac{dy}{dx} e^x -2 \frac{dy}{dx} e^x- x e^y \left(\frac{dy}{dx} \right)^2}{e^x + xe^y}. \end{align*} Note that this is continuous in a neighborhood of $x=0$ as the denominator is non-zero near $(x,y)=(0,0)$ and the numerator is composed of $C^1$ functions.
If $x=0$, then $y=0$ as well due to the original equation. Using our first bit of implicit differentiation, we have that $$\frac{dy}{dx}\bigg|_{(x,y)=(0,0)} = -1.$$ Substituting everything into our expression for $\frac{d^2 y}{dx^2}$ yields $$\phi''(0) = \frac{d^2 y}{dx^2} \bigg|_{(x,y)=(0,0)}= \frac{-0 -4 (-1)e^0 -0}{e^0 + 0} =4. $$