#### Answer

$(a)$ $$\frac{dy}{dx}=\frac{1-y\sec^2(xy)}{x\sec^2(xy)-1}$$
$(b)$ The slope of the tangent line to the curve is $m=-1$

#### Work Step by Step

$(a)$ Use implicit differentiation to determine $\frac{dy}{dx}$ for $\tan xy=x+y$
Taking the derivative implicitly we get (using chain rule for $\tan xy$):
$$\sec^2(xy)\left((1)y+x\frac{dy}{dx}\right)=1+\frac{dy}{dx}$$
solve for $\frac{dy}{dx}$ by getting both $\frac{dy}{dx}$ on the same side
$$\frac{dy}{dx}=\frac{1-y\sec^2(xy)}{x\sec^2(xy)-1}$$
$(b)$ Find the slope of the tangent line to the curve at $(0,0)$
We plug in the point $(0,0)$ into the derivative from part $(a)$. Hence:
$$\frac{dy}{dx}=\frac{1-(0)\sec^2((0)(0))}{(0)\sec^2((0)(0))-1}=\frac{1}{-1}=-1$$