What does $\frac{d}{d(r\sin(\theta)\cos(\phi))}$ equal? I am very confused because if the function were in the numerator, not the denominator, it would make perfect sense to me (i.e., $\frac{d(r\sin(\theta)\cos(\phi))}{dr}$ or $\frac{d(r\sin(\theta)\cos(\phi))}{d\theta}$ or $\frac{d(r\sin(\theta)\cos(\phi))}{d\phi)}$, but since it is $\frac{d}{d(function)}$, I don't know what to do with this. Moreover, I know what the solution is, but no clue how it is obtained.
Solution is apparently $\sin(\theta)\cos(\phi)\frac{d}{dr} + \frac{1}{r}\cos(\theta)\cos(\phi)\frac{d}{d\theta} - \frac{1}{r} \frac{\sin(\phi)}{\sin(\theta)}\frac{d}{d\phi}$
I notice that the solution is similar to $\frac{d(r\sin(\theta)\cos(\phi))}{dr} + \frac{d(r\sin(\theta)\cos(\phi))}{d\theta} + \frac{d(r\sin(\theta)\cos(\phi))}{d\phi}$, but not exactly.
Just slug through the spherical coordinates change of variables, likely in your textbook. Note your derivatives are all partial derivatives, and here you are meant to keep x,y constant!
You know that $$ x= r\sin\theta \cos\phi, $$ so $$ \partial_x= \partial_x r ~\partial_r + \partial_x \theta ~\partial_\theta + \partial_x \phi ~ \partial_\phi \\ =\frac{x}{r} ~\partial_r +\frac{1}{r}\cos\theta\cos\phi ~\partial_\theta -\frac{1}{r}\frac{\sin\phi}{\sin\theta} ~ \partial_\phi~~,$$ your very expression, as $x/r=\sin\theta ~\cos\phi $. Note the inverse Jacobian $J^{-1}$.