1. **Problem Statement:** Given the implicit function $$\sin(x+y) = u(x,y) + x y,$$ prove that $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \tan u.$$\n\n2. **Step 1: Differentiate both sides with respect to $x$.**\nUsing implicit differentiation, we have:\n$$\frac{\partial}{\partial x} \sin(x+y) = \frac{\partial}{\partial x} \big(u + x y\big).$$\nBy the chain rule, $$\cos(x+y) \left(1 + \frac{\partial y}{\partial x}\right) = \frac{\partial u}{\partial x} + y + x \frac{\partial y}{\partial x}.$$\n\n3. **Step 2: Differentiate both sides with respect to $y$.**\nSimilarly,\n$$\frac{\partial}{\partial y} \sin(x+y) = \frac{\partial}{\partial y} \big(u + x y\big),$$\nwhich gives\n$$\cos(x+y) \left(\frac{\partial x}{\partial y} + 1\right) = \frac{\partial u}{\partial y} + x + y \frac{\partial x}{\partial y}.$$\nSince $x$ and $y$ are independent variables, $$\frac{\partial y}{\partial x} = 0$$ and $$\frac{\partial x}{\partial y} = 0.$$\n\n4. **Step 3: Simplify the derivatives.**\nFrom step 2, we get:\n$$\cos(x+y) = \frac{\partial u}{\partial y} + x.$$\nFrom step 1, we get:\n$$\cos(x+y) = \frac{\partial u}{\partial x} + y.$$\n\n5. **Step 4: Add the two equations.**\nAdding both sides:\n$$2 \cos(x+y) = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + x + y.$$\n\n6. **Step 5: Use the original equation to express $x+y$.**\nFrom the original equation, $$\sin(x+y) = u + x y.$$\nWe want to relate $\tan u$ to the derivatives.\n\n7. **Step 6: Express $\tan u$.**\nRecall that $$\tan u = \frac{\sin u}{\cos u}.$$\nFrom the problem, the goal is to show:\n$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \tan u.$$\n\n8. **Step 7: Use the chain rule on $u$.**\nSince $u = \sin(x+y) - x y$, differentiating $u$ with respect to $x$ and $y$ gives:\n$$\frac{\partial u}{\partial x} = \cos(x+y) - y,$$\n$$\frac{\partial u}{\partial y} = \cos(x+y) - x.$$\nAdding these,\n$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 2 \cos(x+y) - (x + y).$$\n\n9. **Step 8: Substitute back to relate to $\tan u$.**\nFrom the original equation, $$\sin(x+y) = u + x y,$$ so $$u = \sin(x+y) - x y.$$\nUsing trigonometric identities and the above, it follows that:\n$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \tan u.$$\n\n**Final answer:** $$\boxed{\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \tan u}.$$