Using implicit differentiation on trigonometric fraction and logarithm I am supposed to solve the following using implicit differentiation.

$$\sin\left(\frac{x}{y}\right)+\ln(y)=xy$$

This is what I have so far:

$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}$$

My problem here is differentiating the fraction as well as the $\ln y$, especially since $\ln y$ does not contain $x$ to differentiate towards.

Thanks for your help in understanding this.
Keep doing what you did on the right-hand side to get

$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}\;.\tag{1}$$

You have

$$\frac{d}{dx}\left(\frac{x}y\right)=\frac{y\cdot1-x\frac{dy}{dx}}{y^2}=\frac1y-\frac{x}{y^2}\frac{dy}{dx}$$

and

$$\frac{d}{dx}\ln y=\frac1y\frac{dy}{dx}\;;$$

both of these are just applications of the chain rule (together with the quotient rule and the rule for differentiating the natural log). Now for convenience I’ll write $y'$ for $dy/dx$; then $(1)$ becomes

$$\cos\left(\frac{x}y\right)\left(\frac1y-\frac{x}{y^2}y'\right)+\frac{y'}y=y+xy'\;,$$

and all that remains is to solve for $y'$.
$$\sin(x/y) + \ln y = xy$$

$$\frac{\cos(x/y)(y-xy')}{y^2} + \frac{y'}{y} =xy' + y$$

$$\frac{y\cos(x/y) - xy'\cos(x/y)}{y^2} + \frac{y'}{y} = (xy' + y)\mid ~~\times ~~y^2$$

we have

$$y\cos(x/y) - xy'\cos(x/y) +  yy' = xy^2y' + y^3$$

$$yy' - xy'\cos(x/y) - xy^2y' =y^3 - y\cos(x/y)$$

$$y'(y-x\cos(x/y)-xy^2)=y^3 - y\cos(x/y)$$  

$$y'= \frac{y^3-y\cos(x/y)}{y-x\cos(x/y)-xy^2}$$