I don't think there is an easy way to do this in general. Gorenstein is a fairly homological / commutative algebraic condition. However, the condition that $K_X$, the canonical divisor, is a Cartier divisor is quite close to the Gorenstein condition and for some purposes, it is just as good.
Another algebraic place to read about Gorenstein singularities (besides Eisenbud's book) include Bruns and Herzog's Cohen-Macaulay rings.
There is also a question you should ask yourself about Gorenstein singularities. Which of the following properties of Gorenstein singularities do you want:
- The fact that Gorenstein singularities are Cohen-Macaulay (and so have well-behaved Serre-duality without the need for fancy homological machinary and derived categories, see the Serre duality section in Hartshorne's Algebraic Geometry).
- The fact that on a Gorenstein variety, the canonical Weil divisor $K_X$ is actually a Cartier divisor.
In fact, a singularity being Gorenstein is equivalent to both conditions 1. and 2. I also think that 1. + 2. is how most geometers think about the Gorenstein condition. Commutative algebraists tend to have a different perspective.
I should also point out perhaps one other large class of rings where you can easily detect whether or not it is Gorenstein (besides the already-mentioned complete intersections).
Suppose that X is a projective variety with an ample line bundle $\mathcal{L}$. Then the section ring:
$$ \oplus_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})$$
is Gorenstein if and only if the following two conditions hold.
$H^i(X, \mathcal{L}^{\otimes n}) = 0$ for all $i > 0$ and all $n \geq 0$. This is just condition 1. above.
$\mathcal{O}_X(K_X)$ is isomorphic to $\mathcal{L}^n$ for some integer $n$. This is condition 2. above.
EDIT: If you have explicit equations, you can often use Macaulay2 to check whether the ring is Gorenstein. Let me know if this would be useful to you.