Implicit function, real analysis homework question Consider the equation $xe^{y}+ye^{x}=0$
(a) Prove that this equation defines $y$ as a $C^{\infty}$ function of $s$ in a neighborhood of $(0,0)$
(b) Ley $y=g(x)$ be this implicitly defined function. Find $g'(0)$ and $g''(0)$
(c) Explain the appearance of the curve $xe^{y}+ye^{x}=0$ near $(0,0)$
I think (a) is the one I need most help with. Really appreciate if someone give me a hint or explain to me how to prove it. 
 A: Set
$F(x, y) = xe^y + ye^x; \tag {1}$
then
$F_y(x, y) = \dfrac{\partial F}{\partial y}(x, y) = xe^y + e^x, \tag{2}$
and evaluating this at $(0, 0)$ we find that
$F_y(0, 0) = 1, \tag{3}$
and now by the implicit function theorem, more about which may be found in the wikipedia page of the same name, we can assert the existence of a smooth function $g(x)$ with $y=g(x)$ in a neighborhood $U$ of $x = 0$, such that
$F(x, y) = xe^y + ye^x = xe^{g(x)} + g(x)e^x = 0 \tag{4}$
for $x \in U$.  This answers part (a).  Note that (4) implies $g(0) = 0$.
For part (b), we differentiate (4) with respect to $x$, yielding
$e^y + xy'e^y + y'e^x + ye^x = 0, \tag{5}$
or
$y'(xe^y + e^x) + e^y + ye^x = 0, \tag{6}$
which may readily be solved for $y'$:
$y' = -\dfrac{e^y + ye^x}{xe^y + e^x}. \tag{7}$
In order to find $y''$, we can in fact differentiate (7) with respect to $x$, but I think it is a little easier to work directly with (6):
$y''(xe^y + e^x) + y'(e^y + xy'e^y + e^x) + y'e^y + y'e^x + ye^x = 0, \tag{8}$
which we re-write in a manner which consolidates the terms containing $y'$:
$y''(xe^y + e^x) + y'(e^y + xy'e^y + e^x + e^y + e^x) + ye^x = 0, \tag{9}$
or
$y''(xe^y + e^x) + y'(2e^y + xy'e^y + 2e^x) + ye^x = 0, \tag{10}$
from which
$y'' = -\dfrac{y'(2e^y + xy'e^y + 2e^x) + ye^x}{xe^y + e^x}. \tag{11}$
Note that all the operations in the above are well-defined in a neighborhood of $(0, 0)$ since
$xe^y + e^x = 1 \ne 0$ at that point.  From (7) we have
$g'(0) = y'(0) = -1, \tag{12}$
and (11) yields
$g''(0) = y''(0) = 4. \tag{13}$
We have thus dispensed with part (b).
As for (c), there's not much to explain, really; it's more a matter of describing, in rough terms, the overall features of the curve $F(x, y) = xe^y + ye^x = 0$ near $(0, 0)$.  With the information we have at hand, this is a straightforward matter, since the function $y = g(x)$ satisfies $F(x, g(x)) = 0$.  Thus we can deduce the properties of the curve $F(x, y) = xe^y + ye^x = 0$ from those we know of $g(x)$.  We see that the curve in question passes through the point $(0, 0)$, and that it has slope $-1$ there, in keeping with the fact that $\nabla F(0, 0) = (1, 1)$, normal to the tangent line $y = -x$ to $y = g(x)$ at $(0, 0)$, and the fact that $x$ and $y$ must be of opposite signs when $(x, y) \ne (0, 0)$.  The fact that $g''(0) = 4 > 0$ indicates that the slope $g'(x)$ is increasing as we move away from $x = 0$, so the curve $F(x, y) = 0$ is apparently concave upward at $(0, 0)$, but apart from a more exhaustive analysis, we can say little else.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
