#sin(xy)=ln(x/y)#. First, let's rewrite that #ln#.

#sin(xy)=ln(x)-ln(y)#

Here is the general case first:

**The derivative with respect to #t#**

#d/dt(sin(xy))=d/dt(ln(x)) - d/dt(ln(y))#

Evaluating the derivatives using the chain rule, we get,

#cos(xy)d/dt(xy) = 1/x d/dt(x) - 1/y d/dt(y)#.

#cos(xy)[dx/dt y + x dy/dt] = 1/x dx/dt - 1/y dy/dt#.

Solve algebraically for whichever one you're interested in.

If you want to find the derivative of #y# with respect to #x#, then you're looking for #dy/dx#. You could replace the #t#'s above with #x#'s, but it's probably more clear if we just start over.

**The derivative with respect to #x#**

#d/dx(sin(xy))=d/dx(ln(x)) - d/dx(ln(y))#

Evaluating the derivatives using the chain rule, we get,

#cos(xy)d/dx(xy) = 1/x - 1/y d/dx(y)#.

#cos(xy)[y + x dy/dx] = 1/x - 1/y dy/dx#.

Before solving for #dy/dx# it is helpful to get rid of the fractions by multiplying both sides of the equation by #xy#.

#xycos(xy)[y + x dy/dx] = y - x dy/dx#.

#xy^2cos(xy)+x^2ycos(xy) dy/dx = y - x dy/dx#.

#x dy/dx + x^2ycos(xy) dy/dx = y - xy^2cos(xy)#

#dy/dx = (y - xy^2cos(xy))/(x+x^2ycos(xy))#