Concernig the modulus: We suppose $a,b \ne 0$. There are two possibilities. Either $f$ has a zero and the minimum of $|f|$ is 0 or $f$ doesn't change signs and the minimum of $|f|$ is the minimum resp. maximum of $f$ if $f$ is negative resp. positive.
In your case, if you write $f$ in the form @Mark suggested $ f(x,y) = a\biggl(x + \frac c{2a}\biggr)^2 + b\biggl(y + \frac{d}{2b}\biggr)^2 + C $ ($C$ to be computed) we have the following cases:
- $a$, $b > 0$ and $C \ge 0$. Then $f \ge 0$ and therefore $|f| = f$ attains its minimum at $(-\frac c{2a}, -\frac d{2b})$.
- $a$, $b < 0$ and $C \le 0$. Then $f \le 0$ and $|f| = -f$ attains its minimum at $(-\frac c{2a}, -\frac d{2b})$.
- $a > 0$, $b < 0$. Then $f$ attains positive and negative values, as for example $f(x,0) \to \infty$, $x \to \infty$ but $f(0,y) \to -\infty$, $y \to \infty$. Therefore by connectedness of $f(\mathbb R^2)$ $f$ has a zero and $|f|$'s minimum is 0.
- $a < 0$, $b > 0$. As in case 3.
- $a, b > 0$, $C < 0$. Then $f(-\frac c{2a}, -\frac d{2b}) = C < 0$ but $f(x,0) \to \infty$, $x \to \infty$. Therefore $f$ has a zero.
- $a,b < 0$, $C > 0$. As in case 5, $f$ has a zero.