Find dy
\[
y=\cot ^{-1}\left(\frac{1}{x^{3}}\right)+\cos ^{-1}(4 x)
\]

Step 1 ：Given the function \(y=\cot ^{-1}\left(\frac{1}{x^{3}}\right)+\cos ^{-1}(4 x)\), we need to find its derivative.

Step 2 ：First, we find the derivative of \(y=\cot ^{-1}\left(\frac{1}{x^{3}}\right)\). The derivative of \(\cot ^{-1}(x)\) is \(-\frac{1}{1+x^2}\), so the derivative of \(\cot ^{-1}\left(\frac{1}{x^{3}}\right)\) is \(-\frac{1}{1+\left(\frac{1}{x^{3}}\right)^2}\) times the derivative of \(\frac{1}{x^{3}}\), which is \(-\frac{3}{x^4}\). Therefore, the derivative of \(y=\cot ^{-1}\left(\frac{1}{x^{3}}\right)\) is \(\frac{3}{x^4(1+x^6)}\).

Step 3 ：Next, we find the derivative of \(y=\cos ^{-1}(4 x)\). The derivative of \(\cos ^{-1}(x)\) is \(-\frac{1}{\sqrt{1-x^2}}\), so the derivative of \(\cos ^{-1}(4 x)\) is \(-\frac{1}{\sqrt{1-(4x)^2}}\) times the derivative of \(4x\), which is \(4\). Therefore, the derivative of \(y=\cos ^{-1}(4 x)\) is \(-\frac{4}{\sqrt{1-16x^2}}\).

Step 4 ：Finally, we add the derivatives of the two parts of the function to find the derivative of the entire function. The derivative of \(y=\cot ^{-1}\left(\frac{1}{x^{3}}\right)+\cos ^{-1}(4 x)\) is \(\frac{3}{x^4(1+x^6)}-\frac{4}{\sqrt{1-16x^2}}\).

Step 5 ：Final Answer: \(\boxed{\frac{3}{x^4(1+x^6)}-\frac{4}{\sqrt{1-16x^2}}}\)