Time-Invariant System
The expression is
$$ y(t) = (3-2j)x(1-t)$$


  1) delay original output y(t) by t0, check resulting expression y(t - t0)  


The first step I suggest is not needed but prevents mixing up variable 't' when the given expression is more complex and helps using the correct sign (it will be + t0 later on); first rewrite the original expression with a different time variable, e.g. u:
$$ y(u) = (3-2j)x(1-u)$$

In order to get the delayed output y(t-t0), you need to substitute \$u=t-t0\$

$$ y(u)|_{u=t-t0} = (3-2j)x(1-u)|_{u=t-t0} $$
$$ y(t-t0) = (3-2j)x(1-(t-t0))$$
$$ y(t-t0) = (3-2j)x(1-t+t0))$$


  2) delay original input x(t) by t0 i.e. x(t - t0), check resulting expression y'(t)  


The expression for y'(t) (the ouput that is generated by the delayed input) is thus:

$$ y'(t) = (3-2j)x_{delayed}(1-t)$$

Now, I suggest to first rewrite the delayed input with a different time variable, so:

$$ x_{delayed}(u) = x(u-t0) $$

and substitute it in y'(t) above:

$$ y'(t) = (3-2j)x_{delayed}(1-t)|_{x_{delayed}(u) = x(u-t0)}$$

$$ y'(t) = (3-2j)x(1-t-t0)$$

Now, you can see that the delayed output \$y(t-t0)\$ is not equal to the output y'(t) generated by a delayed input

$$ y(t-t0) \neq y'(t)$$




So, for the delayed output you rewrite the original expression with a different time variable and substitue (t-t0) for that variable.
(You can also directly substitute t=(t-t0), but doing in steps may prevent errors)  
For the output y'(t) generated by the delayed input x(t-t0) you rewrite the delayed input with a different time variable and substitute this in y'(t)