Your first method was correct. The limits as $x\to-\infty$ or $x\to\infty$ can produce very different results, for instance: $\lim_{x\to\infty} x \ne \lim_{x\to -\infty}x\ .$
Here's how to go about solving this limit: $\lim_{x\to - \infty} \frac {3x^2+3x}{2x^2+2}=\lim_{x\to -\infty}\left(\frac {3x^2+3x}{2x^2+2}\cdot\frac{\frac 1 {x^2}}{\frac 1 {x^2}}\right)=\lim_{x\to - \infty} \frac {3 + \frac 3 x}{2 + \frac 2 {x^2}}=\frac 3 2$ The last inequality follows by noting that:
- The limit of a quotient is the quotient of the limits
- The limit of a sum is the sum of the limits
In general, when you have $x\to\infty$ or $x\to -\infty$ and a rational function, try dividing out the highest degree of $x$ from the numerator and denominator.
In the example above, it did not matter that $x\to -\infty$ rather than $x\to\infty$, because all the terms which were dependent on $x$ approached 0 in the limit. A small change to the example, however, does yield different results: $\lim_{x\to- \infty}\frac {3x^3+3x}{2x^2+2}=\lim_{x\to-\infty}\frac{3x+\frac 3 x}{2+\frac 2 {x^2}}=-\infty$ The last equality comes from the observation that the numerator of this expression will decrease without bound, while the denominator approaches 2. Suppose instead that the limit subscript had been $x\to\infty$. Then: $\lim_{x\to \infty}\frac {3x^3+3x}{2x^2+2}=\lim_{x\to\infty}\frac{3x+\frac 3 x}{2+\frac 2 {x^2}}=\infty$ My point is just that you do not get the same result. One way of handling the $-\infty$ case is to make the substitution you suggest, another way would be to just think (similarly as you do for $x\to\infty$) about what happens to the function as $x$ decreases without bound.