Okay so I have $$f(x)=8\tan^{-1}\left(\frac{y}{x}\right)-\ln \left(\sqrt{x^2+y^2}\right)$$ since $$\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(x)=\frac{1}{1+x^2}$$would $$\frac{\mathrm{d}}{\mathrm{d}x}8\tan^{-1}\left(\frac{y}{x}\right)=8\left(\frac{1}{1+\left(\frac{y}{x}\right)^2}\right)$$ ?? If not, how do I go about finding the derivative? Thanks in advance!!
EDIT: CHANGED NOTATION
Using Chain Rule, $$\frac{d(\tan^{-1}\frac yx)}{dx}=\frac{d(\tan^{-1}\frac yx)}{d\frac yx}\cdot\frac{d\frac yx}{dx}$$
As you have found, $$\frac{d(\tan^{-1}\frac yx)}{d\frac yx}=\frac1{1+\left(\frac yx\right)^2}$$
Using Product Rule,
$$\frac{d\frac yx}{dx}=y\frac{d\frac1x}{dx}+\frac1x\frac{dy}{dx}=-\frac y{x^2}+\frac1x\frac{dy}{dx}$$
Again, $$\ln\sqrt{x^2+y^2}=\ln(x^2+y^2)^{\frac12}=\frac12\ln(x^2+y^2)$$
Again apply Chain Rule