7. Find the first partial derivatives for $f(x, y)=9 x \cos (5 x y)$.
\[
\begin{array}{c}
\frac{\partial f}{\partial x}= \\
\frac{\partial f}{\partial y}=
\end{array}
\]

Step 1 ：The given function is \(f(x, y)=9 x \cos (5 x y)\).

Step 2 ：To find the first partial derivative of the function with respect to \(x\), we differentiate the function with respect to \(x\), treating \(y\) as a constant.

Step 3 ：The first partial derivative of the function with respect to \(x\) is \(-45xy\sin(5xy) + 9\cos(5xy)\).

Step 4 ：To find the first partial derivative of the function with respect to \(y\), we differentiate the function with respect to \(y\), treating \(x\) as a constant.

Step 5 ：The first partial derivative of the function with respect to \(y\) is \(-45x^2\sin(5xy)\).

Step 6 ：So, the first partial derivatives of the function \(f(x, y)=9 x \cos (5 x y)\) are \(\frac{\partial f}{\partial x}= -45xy\sin(5xy) + 9\cos(5xy)\) and \(\frac{\partial f}{\partial y}= -45x^2\sin(5xy)\).

Step 7 ：\(\boxed{\frac{\partial f}{\partial x}= -45xy\sin(5xy) + 9\cos(5xy)}\)

Step 8 ：\(\boxed{\frac{\partial f}{\partial y}= -45x^2\sin(5xy)}\)