10. Find the first partial derivatives of the following function. \[ \begin{array}{l} f(x, y)=\sqrt{4 x^{2}+2 x y+3 y^{2}} \\ f_{x}=\square \\ f_{y}=\square \end{array} \]

Step 1 ：First, we need to find the first partial derivative with respect to x, denoted as \(f_{x}\).

Step 2 ：We use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 3 ：The outer function is \(\sqrt{u}\), and its derivative is \(\frac{1}{2\sqrt{u}}\).

Step 4 ：The inner function is \(4x^{2}+2xy+3y^{2}\), and its derivative with respect to x is \(8x+y\).

Step 5 ：Applying the chain rule, we get \(f_{x} = \frac{1}{2\sqrt{4x^{2}+2xy+3y^{2}}} \cdot (8x+y)\).

Step 6 ：Next, we need to find the first partial derivative with respect to y, denoted as \(f_{y}\).

Step 7 ：The outer function is still \(\sqrt{u}\), and its derivative is still \(\frac{1}{2\sqrt{u}}\).

Step 8 ：The inner function is still \(4x^{2}+2xy+3y^{2}\), but its derivative with respect to y is \(x+6y\).

Step 9 ：Applying the chain rule, we get \(f_{y} = \frac{1}{2\sqrt{4x^{2}+2xy+3y^{2}}} \cdot (x+6y)\).

Step 10 ：So, the first partial derivatives of the function are \(f_{x} = \frac{1}{2\sqrt{4x^{2}+2xy+3y^{2}}} \cdot (8x+y)\) and \(f_{y} = \frac{1}{2\sqrt{4x^{2}+2xy+3y^{2}}} \cdot (x+6y)\).

Step 11 ：Finally, we simplify the expressions to get the final results: \(f_{x} = \frac{4x+y}{\sqrt{4x^{2}+2xy+3y^{2}}}\) and \(f_{y} = \frac{x+3y}{\sqrt{4x^{2}+2xy+3y^{2}}}\).

Step 12 ：\(\boxed{f_{x} = \frac{4x+y}{\sqrt{4x^{2}+2xy+3y^{2}}}, f_{y} = \frac{x+3y}{\sqrt{4x^{2}+2xy+3y^{2}}}\)