Let $f(x, y)=\sqrt{160-2 x^{2}-4 y^{2}}$
For which value(s) of $y$ is $f(-3, y)=10$ ?
\[
y=
\]

Step 1 ：We are given the function \(f(x, y)=\sqrt{160-2 x^{2}-4 y^{2}}\) and we are asked to find the value of \(y\) for which \(f(-3, y)=10\).

Step 2 ：We substitute \(x=-3\) and \(f(-3, y)=10\) into the function to get the equation \(10 = \sqrt{160-2*(-3)^{2}-4 y^{2}}\).

Step 3 ：We square both sides of the equation to get rid of the square root, resulting in \(100 = 160 - 18 - 4y^{2}\).

Step 4 ：We rearrange the equation to solve for \(y\), getting \(4y^{2} = 42\).

Step 5 ：Finally, we solve for \(y\) to get \(y=\pm\frac{\sqrt{42}}{2}\).

Step 6 ：So, the values of \(y\) for which \(f(-3, y)=10\) are \(y=\boxed{-\frac{\sqrt{42}}{2}, \frac{\sqrt{42}}{2}}\).