No, that just says "1= 1" and that doesn't help at all.
There is no point in posting here if you will not at least try what people suggest.
Your original problem is
$\displaystyle \frac{3}{x-1}= \frac{2}{2x+3}$
You then multiply both sides by the product of the denominators (x-1)(2x+3) to get
$\displaystyle \frac{3}{x-1}(x-1)(2x+3)= \frac{2}{2x+3}(x-1)(2x+3)$
No, as tonio told you, the "x- 1" terms cancel on the right and the "2x+3" terms cancel on the left leaving
$\displaystyle 3(2x+3)= 2(x-1)$
which is the same as
$\displaystyle 6x+ 3= 2x- 2$
Can you solve that equation?
You really should ask yourself "Why am I taking the steps that I'm taking ?"
If two fractions are equal, the pattern is:
$\displaystyle \frac{1}{2}=\frac{2}{4}=\frac{3}{6}=\frac{4}{8}=\f rac{5}{10}=\frac{6}{12}=......$
$\displaystyle \frac{3}{5}=\frac{6}{10}=\frac{9}{15}=\frac{12}{20 }=......$
Notice that the denominator of the first fraction "times" the numerator of the second fraction
is equal to the numerator of the first fraction "times" the denominator of the second one.
In other words... $\displaystyle \frac{2}{4}=\frac{3}{6}\Rightarrow\ 2*6=3*4$
$\displaystyle \frac{3}{5}=\frac{6}{10}\Rightarrow\ 3*10=5*6$
and so on.
Even when you have "x" in the fractions, it's still true
$\displaystyle \displaystyle\frac{3}{x-1}=\frac{2}{2x+3}\Rightarrow\ 3(2x+3)=2(x-1)$
as Wilmer explained.
If you want to play with common denominators as you were doing,
it's more work, but you need to choose one of a few ways to continue.
One way is to say "2 things are equal so we get zero by subtracting them".
$\displaystyle \displaystyle\frac{3}{x-1}=\frac{2}{2x+3}\Rightarrow\frac{3}{x-1}-\frac{2}{2x+3}=0$
Now you can do your "cross-multiplication stuff"
$\displaystyle \displaystyle\frac{3(2x+3)-2(x-1)}{(x-1)(2x+3)}=0$
which means the numerator is zero
$\displaystyle 3(2x+3)-2(x-1)=0\Rightarrow\ 6x+9-2x+2=0$
Of course, cross multiplication is just the following:
$\displaystyle \displaystyle\frac{3}{x-1}-\frac{2}{2x+3}=0\Rightarrow\frac{3}{x-1}\left[\frac{2x+3}{2x+3}\right]-\frac{2}{2x+3}\left[\frac{x-1}{x-1}\right]=0$
and again the numerator is zero.
Another way is to not bother subtracting at all:
$\displaystyle \displaystyle\frac{3}{x-1}=\frac{2}{2x+3}\Rightarrow\frac{3}{x-1}\left[\frac{2x+3}{2x+3}\right]=\frac{2}{2x+3}\left[\frac{x-1}{x-1}\right]$
and again, since the denominators are equal, the numerators must also be equal,
so we again obtain
$\displaystyle 3(2x+3)=2(x-1)$
Your working is correct here, but be careful with the notation! Look where the equal signs are below:
$\displaystyle 6x+9=2x-2$
$\displaystyle 4x = -11$
$\displaystyle x = -\frac{11}{4}$
$\displaystyle =-2.75$
I know it's picky, but in certain exams, method marks may not be granted if your equations are not set out correctly, regardless of whether your solution is correct, so it's best to be as clear as possible. Obviously, here it's obvious what you're trying to communicate, but it's best to avoid getting into any nasty habits.