Before we start, I'd like to simplify the given equation.
#9 = e^(y-x) - xy#, this may make things easier.
First step in implicit differentiation is to apply the derivative operator to both sides.
#d/dx [9] = d/dx [e^(y-x) - xy]#, the left hand side is trivial, namely, it's just #0#. By applying the chain rule to #e^(y-x)# and the product rule to #xy#, we obtain the following:
#0 = e^(y-x) * (dy/dx - 1) - (1 * y - x dy/dx)#
#0 = e^(y-x) dy/dx - e^(y-x) - y + x dy/dx#
Doing a bit of algebra, and factorising that #dy/dx#, we get,
#e^(y-x) + y = (e^(y-x) + x) dy/dx#
Therefore, #dy/dx = (e^(y-x) + y)/(e^(y-x) + x)#.
I suppose, to simplify further, we can multiply the numerator and the denominator by #e^x# on the right hand side:
#dy/dx = (e^y + ye^x)/(e^y + xe^x)#.
