Find the first partial derivatives of the following function.
\[
\begin{array}{l}
\quad F(u, v, w)=\frac{u}{v+6 w} \\
F_{u}=\square \\
F_{v}=\square \\
F_{w}=\square
\end{array}
\]

Step 1 ：The function given is \(F(u, v, w) = \frac{u}{v+6w}\).

Step 2 ：We are asked to find the first partial derivatives of this function with respect to u, v, and w.

Step 3 ：Partial derivatives are a type of derivative where we differentiate with respect to one variable while keeping the other variables constant.

Step 4 ：To find the partial derivative with respect to u, we treat v and w as constants and differentiate normally. This gives us \(F_u = \frac{1}{v+6w}\).

Step 5 ：To find the partial derivative with respect to v, we treat u and w as constants. This is a bit more complex because v is in the denominator of a fraction. We'll need to use the quotient rule, which states that the derivative of a function in the form of f/g is \((f'g - fg')/g^2\). This gives us \(F_v = -\frac{u}{(v+6w)^2}\).

Step 6 ：Similarly, to find the partial derivative with respect to w, we treat u and v as constants and again use the quotient rule because w is in the denominator of a fraction. This gives us \(F_w = -\frac{6u}{(v+6w)^2}\).

Step 7 ：So, the first partial derivatives of the function are: \(F_u = \frac{1}{v+6w}\), \(F_v = -\frac{u}{(v+6w)^2}\), and \(F_w = -\frac{6u}{(v+6w)^2}\).

Step 8 ：\(\boxed{F_u = \frac{1}{v+6w}, F_v = -\frac{u}{(v+6w)^2}, F_w = -\frac{6u}{(v+6w)^2}}\)