Intuition for multivariable functions Normal equations are easy enough to perceive as geometrical shapes or volumes when defined explicitly in the form $y=f(x)$.
But how do I understand equations like $g(x+y)=e^yg(x)+e^xg(y)$? I can make some sense of it algebraically, with it being a definition of a function and $x$ and $y$ being two variables. But how is it intuitively represented?
 A: A hint:
The functional equation in your question reminded me of the following: The exponential function satisfies the functional equation $e^{x+y}=e^x\cdot e^y\>$.  Therefore I would bring this into the picture by trying the Ansatz
$$g(x):= e^x\cdot h(x)$$
with a new unknown function $h$. Maybe a simpler problem for $h$ results.
A: $\def\e{\varepsilon} 
\def\To{\rightarrow}$Note that $g$ in
\begin{align*}
g(x+y) &= e^y g(x)+e^x g(y) \tag{1} 
\end{align*}
is not a multivariable function.
It is a function of one variable.
This is an equation (called a functional equation) for which the unknown is a function, namely the function $g$.
That is, to solve this equation involves not finding an $x$- or $y$-value for which the equation is satisfied but to find a function for which the equation holds for all $x$, $y$.
The geometric meaning of (1) is clear:
the value of $g$ at $x+y$ is related to its value at the points $x$ and $y$ as well as on $x$ and $y$ directly through (1).
Let us examine a simpler relationship implied by (1),
\begin{align*}
g(2x) = 2e^x g(x).\tag{2}
\end{align*}
This states that the height of the function at $2x$ is $2e^x$ times the height of the function at $x$.
(Let $x=1$ so
$g(2) = 2e g(1)$.
This states that the height of the function $g$ at 2 is $2e$ times the height of the function at 1.)
A relationship such as (1) is extremely restrictive and almost completely specifies the function $g$.
A good intuition comes from recurrence relations.
Suppose that
\begin{align*}
f(0) &= a \\
f(n+1) &= r f(n)
\end{align*}
where $n$ is a nonnegative integer.
This is the recurrence relation for the geometric sequence.
Note the similarity with (1):
the value of $f$ at $n+1$ is related to its value at $n$.
The geometric sequence could instead be defined by
\begin{align*}
f(0) &= a \\
f(m+n) &= r^m f(n),
\end{align*}
where $m,n$ are nonnegative integers, in which case the similarity to (1) is even more transparent.
Addendum
Let us agree that we wish to solve (1) on the reals.
Let $x=y=0$.
Then
$g(0) = 2g(0)$.
Thus, $g(0) = 0$.
Consider
\begin{align*}
g(x+\e) - g(x)
    &= (e^\e-1)g(x) + e^x g(\e). \tag{3}
\end{align*}
This implies that if $\lim_{\e\To0}g(\e)=0$, that is, if $g$ is continuous at $x=0$, that $g$ is continuous for all reals.
In fact, if we assume that $g$ is differentiable at $x=0$ this implies that $g$ is differentiable for all reals since
\begin{align*}
g'(x) &= \lim_{\e\To0} \frac{g(x+\e)-g(x)}{\e} 
    = g(x) + e^x g'(0). \tag{4}
\end{align*}
We can begin graphing $g$ by point plotting using (2).
Suppose, without loss of generality, that $g(1) = e$.
Since $g(2x) = 2e^x g(x)$,
\begin{align*}
g(2) &= 2e^1g(1) 
    = 2e^2 \\
g(4) &= 4e^4 \\
\vdots
\end{align*}
Similarly, since $g(x/2) = e^{-x/2}g(x)/2$,
\begin{align*}
g(1/2) &= \frac{e^{-1/2}}{2}g(1) 
    = \frac{e^{1/2}}{2} \\
g(1/4) &= \frac{e^{1/4}}{4} \\ 
\vdots
\end{align*}
See the figures below.
(Values of $g$ for $x<0$ may be found by setting $y=-x$ in (1), with the result
$g(-x) = -e^{-2x}g(x)$.)
How to find the form of $g$ for a generic $x$-value?
There are many ways to approach this.
Here we solve the differential equation (4).
Applying the condition that $g(0) = 0$ we find the solution
$g(x) = x e^x g'(0).$
If $g(1) = e$ we find
$$g(x) = x e^x.$$
(Note that this is equivalent to the condition that $g'(0) = 1$.)
The graph of this function is given in the figures below as a solid line.
(One can also transform this problem as indicated by @Christian Blatter. This will result in a well-known functional equation.
It should be noted that the conditions that $g$ be continuous or differentiable at $0$ are not necessary and that generally the solution may be quite pathological.)

Figure 1. Plot of $g(2^k)$ for $k=0,1,\ldots$.

Figure 2. Plot of $g(2^{-k})$ for $k=0,1,\ldots$.
