Find $d y$ \[ y=\frac{7 x}{1+3 x^{2}} \]

Step 1 ：Define the functions \(f(x) = 7x\) and \(g(x) = 1 + 3x^2\).

Step 2 ：Calculate the derivatives of these functions, \(f'(x) = 7\) and \(g'(x) = 6x\).

Step 3 ：Use the quotient rule to find the derivative of \(y\), which is \(dy = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\).

Step 4 ：Substitute the functions and their derivatives into the formula to get \(dy = \frac{7(1 + 3x^2) - 7x(6x)}{(1 + 3x^2)^2}\).

Step 5 ：Simplify the expression to get the final answer: \(dy = \frac{7 - 21x^2}{(1+3x^2)^2}\).

Step 6 ：The derivative of the function \(y=\frac{7 x}{1+3 x^{2}}\) is \(\boxed{\frac{7 - 21x^2}{(1+3x^2)^2}}\).