Gauss law is sometimes used to find the electric field, but this is usually confined to situations where there is a strong symmetry that allows you to conclude that the electric field you're looking for is precisely the one in Gauss law, and moreover it reduces to something easy to evaluate, like the product of the magnitude of $\mathbf E$ with a surface. Apart from these cases, Gauss law simply relates the flux of the total electric field through a closed surface and the total charge contained within it. For (essentially) geometrical reasons, outside charges do not contribute to the flux, but they indeed contribute to the total electric field.
The answer to your question involves the fact that one does not usually know a priori the electric field $\textbf{E}$ (or, for that matter, its direction) of a charge distribution $\rho$.
Gauss's law, in integral form, relates the flux of the electric field through some closed surface $S$ to the charge enclosed within the volume bounded by $S$. Precisely, it is the statement that given an electric field $\textbf{E}(\textbf{r})$ defined over space, the flux integral over any closed surface $S$ will always yield
$$ \oint_S \textbf{E} \cdot d\textbf{a} = \frac{Q_\text{enc}}{\epsilon_0}.$$
Normally surface integrals over vector fields involve parametrizing the surface (i.e. describing a two-dimensional surface by two parameters $u,v$ related to the Euclidean coordinates $x,y,z$). Even then, one has to additionally compute the product
$$\textbf{E} \cdot d\textbf{a} = \textbf{E} \cdot \hat{n}da,$$
where $\hat{n}$ is the unit normal to the surface and can be calculated from the parametrization. This quantity can assume different values everywhere along the surface.
So far I've only talked about the difficulties in computing the flux integral of a vector field over a general surface. When using Gauss's law, we have the added problem of not knowing the electric field (this is the quantity we're trying to find!). We now are tasked with computing an integral over an undefined function! This is where symmetry comes in and saves the troubled physicist.
Essentially, symmetric charge distributions allow one to choose a convenient surface (which preserves the symmetry) to remove $\textbf{E}$ from the integral. For example, consider a uniformly charged spherical volume of radius $R$ (i.e. a ball). Due to symmetry, one can argue that the electric field generated from this distribution must be radially symmetric. If we take our surface $S$ to be a sphere of radius $r$, then we find that the normal to the sphere and the direction of the electric field coincide, so
$$\textbf{E} \cdot d\textbf{a} = |\textbf{E}|\oint_S da = 4 \pi r^2 |\textbf{E}|,$$
since we are now simply computing the surface area of a sphere. We can now simply divide to find the answer:
$$ |\textbf{E}| = \frac{1}{4 \pi \epsilon_0} \frac{Q_\text{enc}}{r^2}.$$
To summarize, Gauss's (integral) law relates the flux integral of the electric field to the charge contained within a surface. Because we do not know the electric field, Gauss's law is only useful when we can remove the electric field from within the integral, which happens when the charge distribution displays certain spatial symmetries (spherical, cylindrical, planar).
Best Answer
Lets suppose that you have a complicated distribution of charges such that there is no symmetry in the distribution. So, basically our Electric field $\vec E$ will also be a complicated function. Now, Gauss' Law states :
$$\oint_S \vec E\cdot \vec {ds} = \frac{q_{enc}}{\epsilon_0}$$
If there is no symmetry in $\vec E$, we cannot easily integrate the left hand side, and thus Gauss' Law can't help us.
So basically, it is applied in cases where the Electric field is somehow constant so that it can be pulled out of the integration.