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?
On
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.
$\def\e{\varepsilon} \def\To{\rightarrow}$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.)