For which value of $C(f)$ would the following inequality hold? I am wondering what would be the value of $C(f)$ for the following inequality to hold? E.g., $C(f)$ could be some quantity related to the Lipschitz constant or the size of the domain.
$$\left(\int f(x,x) dx\right)^2 \leq C(f)\cdot \iint f(x,y)^2 dxdy$$
 A: The question is not well posed, as it is not quite clear in what terms you want $C(f)$ to be expressed.
If one only uses the terms you did mention -- the Lipschitz constant and the size of the domain, then no finite $C(f)$ exists. Indeed, suppose that the domain, of "size" $a\in(0,\infty)$, is $[0,a]^2$. Take any real $c>0$ and let
\begin{equation*}
    f(x,y):=\max(0,c-|y-x|)
\end{equation*}
for real $x$ and $y$. Then $f$ is $1$-Lipschitz, the left-hand side of your inequality is
\begin{equation*}
    \text{lhs}:=\Big(\int f(x,x) dx\Big)^2=c^2a^2,
\end{equation*}
and the double integral on the right-hand side of your inequality is
\begin{equation*}
\begin{aligned}
    \text{rhs}&:=\iint f(x,y)^2\,dx\,dy \\ 
    &\le\int_0^a dx\,\int_{-\infty}^\infty dy\,\max(0,c-|y-x|)^2 \\ 
    &=\int_0^a dx\,\int_{x-c}^{{x+c}} dy\,(c-|y-x|)^2  \\ 
    &=2\int_0^a dx\,\int_0^c dz\,(c-z)^2\le c^3a. 
\end{aligned}
\end{equation*}
So, if your inequality holds, then
\begin{equation*}
    C(f)\ge\frac{\text{lhs}}{\text{rhs}}\ge\frac{c^2a^2}{c^3a}=\frac ac. 
\end{equation*}
Since the "height" $c$ of the function $f$ can be arbitrarily small, there is no finite $C(f)$ expressed in terms of the Lipschitz constant and the size of the domain such that your inequality holds.

On a somewhat positive note, suppose that $f$ is $L$-Lipschitz for some $L\in(0,\infty)$, so that $|f(x,y)|\ge\max(0,|f(x,x)|-L|y-x|)$ for all $x,y$. Then
\begin{equation*}
\begin{aligned}
    \iint f(x,y)^2\,dx\,dy &\ge\int dx\,\int dy\,\max(0,|f(x,x)|-L|y-x|)^2 \\ 
    &=\frac23\,\int dx\,\frac{|f(x,x)|^3}L  \\ 
    &=\frac2{3L}\,\int dx\,|f(x,x)|^3. 
\end{aligned}
\tag{5}\label{5}
\end{equation*}
Remark: We see that, perhaps unexpectedly, the third power of $f$ appears here. However, the "physical" dimensions of $\iint f(x,y)^2\,dx\,dy$ and $\frac2{3L}\,\int dx\,|f(x,x)|^3$ are the same: If, say, we measure $f$ in grams (g) and $x,y$ in centimeters (cm), then $\iint f(x,y)^2\,dx\,dy$ and $\frac2{3L}\,\int dx\,|f(x,x)|^3$ will both be measured in g$^2\,$cm$^2$, because $L$ will be measured in g/cm. This confirms that the third power of $f$ is the right one in $\int dx\,|f(x,x)|^3$.
(I believe that the centimeter-gram-second system of units (CGS) -- with its three base units -- is much better than the (unfortunately) universally accepted International System of Units (Si) with its seven base units and a myriad derived units. :-))
Assuming that the "size" of the domain understood as the Lebesgue measure $|D|$ of the set $D:=\{x\colon f(x,x)\ne0\}$ is finite, by the Hölder inequality we get
\begin{equation*}
    \int dx\,|f(x,x)|^3=\int_D dx\,|f(x,x)|^3\ge\frac1{|D|^2}\,\Big(\int_D dx\,|f(x,x)|\Big)^3. 
\end{equation*}
So,
\begin{equation*}
    \Big(\int dx\,f(x,x)\Big)^2\le\Big(\int dx\,|f(x,x)|\Big)^2 \\ 
    \le\Big(\frac{3L|D|^2}2\Big)^{2/3}
    \Big(\iint f(x,y)^2\,dx\,dy\Big)^{2/3}. \tag{10}\label{10}
\end{equation*}
We saw in the first, "negative" part of this answer that the reason why there is no finite $C(f)$ expressed in terms of the Lipschitz constant and the size of the domain such that your inequality holds is that the "height" of the function $f$ can be arbitrarily small. Now, if we try to avoid this by requiring that, say,
\begin{equation*}
    \iint f(x,y)^2\,dx\,dy\ge b^2>0,
\end{equation*}
then \eqref{10} will yield
\begin{equation*}
    \Big(\int dx\,f(x,x)\Big)^2
    \le\Big(\frac{3L|D|^2}{2b^2}\Big)^{2/3}\,
\iint f(x,y)^2\,dx\,dy. 
\end{equation*}
Alternatively, if we try to avoid small $f$'s by requiring that
\begin{equation*}
    \int dx\,|f(x,x)|\ge h>0,
\end{equation*}
then \eqref{10} will yield
\begin{equation*}
    \Big(\int dx\,f(x,x)\Big)^2
    \le\frac{3L|D|^2}{2h}\,
\iint f(x,y)^2\,dx\,dy. 
\end{equation*}
