$xe^y+ye^x=0$ solution is second differentiable 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. 
 A: 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
$$\phi''(0) = \frac{d^2 y}{dx^2} \bigg|_{(x,y)=(0,0)}= \frac{-0 -4 (-1)e^0 -0}{e^0 + 0} =4.   $$
