Update 4: I found the following, updated integral identity:
$$\int_{l=-\infty}^\infty l \left. \left( \frac{\partial}{\partial x} f(x,y)\right) \right\vert_{x=y=l} \mathrm{d} l = -\int_{l=-\infty}^\infty l \left. \left( \frac{\partial}{\partial y} f(x,y)\right) \right\vert_{x=y=l} \mathrm{d} l - \int_{l=-\infty}^\infty f(l,l) \mathrm{d} l.$$
This assumes that $f$, $f_x$ and $f_y$ tend to $0$ sufficiently fast. In short, it reads $$\int_{l=-\infty}^\infty l f_x(l, l) \mathrm{d} l= - \int_{l=-\infty}^\infty l f_y(l,l) \mathrm{d} l - \int_{l=-\infty}^\infty f(l,l) \mathrm{d} l,$$
or, even shorter, $$\int_{l=-\infty}^\infty (f + lf_x + lf_y)(l,l) \mathrm{d}l = 0.$$
I wonder how this equation can be proven using integration by parts or substitution, for two reasons: I want to learn how it works, and I want to double-check the derivation of the equation shown below.
Update 1: This is how I derived the equation. Start with
$$F(m,n) := \int_{l=-\infty}^{\infty} f(ml, nl) \mathrm{d} l.$$
For simplicity, assume $m,n>0$. Then
$$F + m \frac{\partial}{\partial m} F + n \frac{\partial}{\partial n} F = F + m F_m + n F_n = 0.$$
(See a proof in Update 3.) Then insert the first equation into the second:
$$ \int_{l=-\infty}^{\infty} f(ml, nl) \mathrm{d} l + m \int_{l=-\infty}^{\infty} \frac{\partial}{\partial m} f(ml, nl) \mathrm{d} l + n \int_{l=-\infty}^{\infty} \frac{\partial}{\partial n} f(ml, nl) \mathrm{d} l \\ = \int_{l=-\infty}^{\infty} f(ml, nl) \mathrm{d} l + m \int_{l=-\infty}^{\infty} l f_x(ml, nl) \mathrm{d} l + n \int_{l=-\infty}^{\infty} l f_y(ml, nl) \mathrm{d} l = 0.$$
Set $m = n = 1$, voilà.
Update 3:
How did I find $F + m F_m + n F_n = 0$?
Start with $$F(m,n+\Delta n) = \int_{l=-\infty}^{\infty} f(ml, (n + \Delta n)l) \mathrm{d} l.$$
Substitute $x =\frac{n + \Delta n}{n}l$, which implies $l =\frac{n}{n + \Delta n}x$ and $\mathrm{d} l =\frac{n}{n + \Delta n} \mathrm{d} x.$
Then $$F(m,n+\Delta n) = \frac{n}{n + \Delta n} \int_{x=-\infty}^{\infty} f(m\frac{n}{n + \Delta n} x, nx) \mathrm{d} x = \frac{n}{n + \Delta n} F(m\frac{n}{n + \Delta n},n).$$
On both sides, subtract $\frac{n}{n + \Delta n} F(m,n)$, divide by $\Delta n$, and take $\lim_{\Delta n \to 0}$. In both cases, we will use that $\frac{n}{n + \Delta n} = 1 - \frac{\Delta n}{n + \Delta n}$.
The left side gives $$\lim_{\Delta n \to 0} \frac{F(m,n+\Delta n) - \frac{n}{n + \Delta n} F(m,n)}{\Delta n} \\ = \lim_{\Delta n \to 0} \frac{F(m,n+\Delta n) - F(m,n) + \frac{\Delta n}{n+\Delta n}F(m,n)}{\Delta n} \\ = \lim_{\Delta n \to 0} \frac{F(m,n+\Delta n) - F(m,n)}{\Delta n} + \frac{\frac{\Delta n}{n+\Delta n}F(m,n)}{\Delta n} \\ = F_n(m,n) + \frac{F(m,n)}{n}.$$
The right side gives $$\lim_{\Delta n \to 0} \frac{\frac{n}{n + \Delta n} F(m\frac{n}{n + \Delta n},n) - \frac{n}{n + \Delta n} F(m,n)}{\Delta n} \\=\lim_{\Delta n \to 0} \frac{n}{n + \Delta n} \frac{ F(m\frac{n}{n + \Delta n},n) - F(m,n)}{\Delta n}.$$
This is $$=\lim_{\Delta n \to 0} \frac{n}{n + \Delta n} \frac{ F(m - \frac{m \Delta n}{n + \Delta n},n) - F(m,n)}{\Delta n}.$$ and further, by expanding with $\frac{m}{n + \Delta n}$, $$=\lim_{\Delta n \to 0} \frac{n}{n + \Delta n} \frac{m}{n + \Delta n} \frac{ F(m - \frac{m \Delta n}{n + \Delta n},n) - F(m,n)}{\frac{m \Delta n}{n + \Delta n}}.$$
We now define $\Delta m := -\frac{m \Delta n}{n + \Delta n}$ to find $$=\lim_{\Delta n \to 0} \frac{n}{n + \Delta n} \frac{m}{n + \Delta n} \cdot \lim_{\Delta m \to 0} \frac{ F(m + \Delta m,n) - F(m,n)}{-\Delta m} = -\frac{m}{n} F_m(m,n).$$
Hence, $F_n(m,n) + \frac{F(m,n)}{n} = -\frac{m}{n} F_m(m,n)$ and thus $$F+mF_m+nF_n=0.$$
This is actually just integration by parts: or rather, let us apply the product rule to $$ z f(z,z). $$ We find: $$ \frac{d}{dz} (zf(z,z)) = f(z,z) + z \frac{d}{dz} f(z,z) = f(z,z) + z\partial_1 f(z,z) + z\partial_2 f(z,z), $$ where $\partial_i$ means the derivative with respect to the $i$th argument of $f$, or more explicitly, $$ \partial_1 f(a,b) = \lim_{h \to 0} \frac{f(a+h,b)-f(a,b)}{h} $$ (this gets around our weird Leibnizian habit of always giving every differentiation variable a name). So in fact, $$ \int_{-\infty}^{\infty} (1+z\partial_1+z\partial_2)f(z,z) \, dz = [z f(z,z)]_{-\infty}^{\infty} = 0, $$ if $f$ decays faster than $z^{-1}$.