If exy = 3, then what is dy/dx at the point (1, ln3)?

If exy=3, then dy/dx will be found as follows:
differentiating our expression implicitly we get:
(x(dy/dx)+y)e^(xy)=0
N/B
when differentiating e^(xy) we apply chain rule. When differentiating xy, we use product rule. 
next, we substitute (1, ln(3)) and solve for dy/dx
(1(dy/dx)+ln(3))^(1ln(3))=0
((dy/dx)+ln(3))e^(ln(3))=0
but
e^(ln3)=3, this is because e and ln  are inverse of each other.

3((dy/dx)+ln(3))=0
dy/dx+ln (3)=0
dy/dx=-ln(3)
The answer is :
dy/dx=-ln(3)

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