Did I calculate the Image of this matrix correctly? 
Calculate the image of the matrix $A= \begin{pmatrix} 1  &  2 & -1\\ 
-2 & -6 & 1 \\  1  & -2 & 0  \end{pmatrix}$

First I transposed the matrix:
$A^{T}= \begin{pmatrix}
1  &  -2 &  1 \\ 
2  &  -6 & -2 \\ 
-1 &   1 &  0 
\end{pmatrix}$
Then I used Gauss to get zero lines. Multiply first line  with $2$, add that to second line:
$A^{T}= \begin{pmatrix}
2  &  -4 & 2\\ 
4  & -10 & 0 \\ 
-1 &   1 & 0 
\end{pmatrix}$
Now multiply third line with $10$ and add second line to third line:
$A^{T}= \begin{pmatrix}
2  &  -4 & 2\\ 
4  & -10 & 0 \\ 
-6 &   0 & 0 
\end{pmatrix}$
We cannot create more zeroes, so Gauss ends here. We transpose back:
$A= \begin{pmatrix}
 2  &   4 & -6 \\ 
-4  & -10 &  0 \\ 
 2  &   0 &  0 
\end{pmatrix}$
Thus we have $\text{Img(A)}= \left\{ \begin{pmatrix}
2\\ 
-4\\ 
2
\end{pmatrix}, \begin{pmatrix}
4\\ 
-10\\ 
0
\end{pmatrix}, \begin{pmatrix}
-6\\ 
0\\ 
0
\end{pmatrix} \right\}$

It's very important for me to know, did I do it correctly? I would do it like this in my exam.

This is a highly unusual method.  However, it is technically correct.  You have indeed found a basis of the image.  Remember to say that the image is the span of those vectors.
