There is no general technique, but there are some simple principles. One is to study the tail behavior of $f$ by comparing it to tractable functions.
By definition, the expectation is the double limit (as $y$ and $z$ vary independently)
$$E_{y,z}[f] = \lim_{y\to-\infty,z\to\infty}\int_y^z x f(x) dx = \lim_{y\to-\infty}\int_y^0 x f(x) dx+ \lim_{z\to\infty}\int_0^z x f(x) dx.$$
The treatment of the two integrals at the right is the same, so let's focus on the positive one. One behavior of $f$ that assures a limiting value is to compare it to the power $x^{-p}$. Suppose $p$ is a number for which $$\liminf_{x\to\infty} x^p f(x)\gt 0.$$ This means there exists an $\epsilon\gt 0$ and an $N\gt 1$ for which $x^p f(x) \ge \epsilon$ whenever $x\in[N,\infty)$. We may exploit this inequality by breaking the integration into the regions where $x\lt N$ and $x \ge N$ and applying it in the second region:
$$\eqalign{
\int_0^z x f(x) dx &=\int_0^{N} x f(x) dx + \int_{N}^z x f(x) dx \\
&=\int_0^{N} x f(x) dx + \int_{N}^z x^{1-p} \left(x^p f(x)\right) dx \\
&\ge \int_0^{N} x f(x) dx + \int_{N}^z x^{1-p} \left(\epsilon\right) dx \\
&= \int_0^{N} x f(x) dx + \frac{\epsilon}{2-p}\left(z^{2-p} - {N}^{2-p}\right).
}$$
Provided $p\lt 2$, the right hand side diverges as $z\to\infty$. When $p=2$ the integral evaluates to the logarithm,
$$\int_{N}^z x^{1-2} \left(\epsilon\right) dx = \epsilon \left(\log(z) - \log(N)\right),$$
which also diverges.
Comparable analysis shows that if $|x|^pf(x)\to 0$ for $p\gt 2$, then $E[X]$ exists. Similarly we may test whether any moment of $X$ exists: for $\alpha\gt 0$, the expectation of $|X|^\alpha$ exists when $|x|^{p+\alpha}f(x)\to 0$ for some $p\gt 1$ and does not exist when $\liminf |x|^{p+\alpha}f(x)\gt 0$ for some $p \le 1$. This addresses the "general question."
Let's apply this insight to the question. By inspection it is clear that $a(x)\approx |x|/\sigma_1$ for large $|x|$. In evaluating $f$, we may therefore drop any additive terms that will eventually be swamped by $|x|$. Thus, up to a nonzero constant, for $x\gt 0$
$$f(x) \approx \frac{\mu_1 x}{\sigma_2 x^3}\phi\left(\frac{\mu_2 x}{\sigma_2 x}\right) = x^{-2}\frac{\mu_1}{\sigma_2}\exp\left(\left(-\frac{\mu_2}{2\sigma_2}\right)^2\right).$$
Thus $x^2 f(x)$ approaches a nonzero constant. By the preceding result, the expectation diverges.
Since $2$ is the smallest value of $p$ that works in this argument--$|x|^pf(x)$ will go to zero as $|x|\to\infty$ for any $p\lt 2$--it is clear (and a more detailed analysis of $f$ will confirm) that the rate of divergence is logarithmic. That is, for large $|y|$ and $|z|$, $E_{y,z}[f]$ can be closely approximated by a linear combination of $\log(|y|)$ and $\log(|z|)$.
