Let $f(x, y)=3^{x}+4 x y$, find $f(0,-3), f(-3,2)$, and $f(3,2)$.

Step 1 ：Given the function \(f(x, y)=3^{x}+4 x y\), we are asked to find the values of the function at three different points: \((0,-3)\), \((-3,2)\), and \((3,2)\).

Step 2 ：To find these values, we substitute the given values of \(x\) and \(y\) into the function and calculate the result.

Step 3 ：For \(f(0,-3)\), we substitute \(x = 0\) and \(y = -3\) into the function to get \(f(0,-3) = 3^{0} + 4*0*(-3) = 1\).

Step 4 ：For \(f(-3,2)\), we substitute \(x = -3\) and \(y = 2\) into the function to get \(f(-3,2) = 3^{-3} + 4*(-3)*2 = -21\).

Step 5 ：For \(f(3,2)\), we substitute \(x = 3\) and \(y = 2\) into the function to get \(f(3,2) = 3^{3} + 4*3*2 = 35\).

Step 6 ：So, the final answers are \(\boxed{f(0, -3) = 1}\), \(\boxed{f(-3, 2) = -21}\), and \(\boxed{f(3, 2) = 35}\).