Deriving 2-D discrete Fourier transforms I have a problem in DFT. It was one of my past-year exam papers questions.

Question: 


  Let $F(u,v)$ be the 2-D Fourier transform of a 2-D continuous function
  $f(x,y)$. Derive in terms of $F(:,:)$ the 2-D Fourier transform of
  each of the following functions
  
  1) $f(x,-2y)$
  
  2) $f(x+2y,y)$


I know how to do 1-D Fourier transforms but not 2-D. I'm not sure how to start on it and need some guidance.



For the second part, this was my approach. Please let me know if it's right or correct me if it's wrong.

Let $τ= x + 2y$ hence $x = τ-2y$ and $dx = dτ$
$$
\begin{align}
\mathfrak{F}\{f(x+2y,y)\}&amp;=∬ f(τ,y)e^{−j2π(u(τ-2y) +vy )} dx\ dv\\ 

\mathfrak{F}\{f(x+2y,y)\}&amp;= ∬ f(τ,y)e^{−j2π(uτ+(-2u+v)y  )} dx\ dτ \\
\mathfrak{F}\{f(x+2y,y)\}&amp;= F(u,-2u+v) 
\end{align}
$$
Here is the first one:

By definition,

$\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,-2y)e^{-j2\pi (ux+vy)}dxdy$

let $\tau = -2y$ and conversely $y=\frac{-\tau}{2}$

$\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd(-\frac{\tau}{2})$

$\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2}\iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd\tau$

$\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2} F(u,-\frac{v}{2})$

The other one will be harder, but I will leave it to you.