Question: This is problem #3 in section 6.1. of the current edition of Stewart's calculus text.

So it's been a while since I've done Calc. II material. I am tempted to try double integrals with this question, but of course, double integrals are not covered in Calc. II. 

Here is my proposed solution:

$$\int\limits\_{-1}^{1}\int\limits\_{y^2-2}^{\exp(y)}\text{ d}x\text{ d}y = \int\limits\_{-1}^{1}\left(e^{y}-y^{2}+2\right)\text{ d}y = \left.e^{y}-\dfrac{y^3}{3}+2y\right|^{y=1}\_{y=-1} = e-\dfrac{1}{3}+2-e^{-1}-\dfrac{1}{3}+2=e-e^{-1}+4-\dfrac{2}{3} \approx 5.684\text{.}$$ 

According to Stewart, this answer is correct. I however, do not have the detailed solutions manual. Is there a way to do this problem WITHOUT double integration, in case I were to teach this to someone in Calc. II?

Answer: 
It seems like you should just compute the relevant partials and then verify that LHS $=$ RHS.

Something like:

$$\frac{\partial u}{\partial x} = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2x \\
\frac{\partial u}{\partial y} = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2y \\
\frac{\partial u}{\partial z} = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2z \\$$ 

The LHS then becomes 

$$\frac{\sin(2x)\sec^2x + \sin(2y)\sec^2y+ \sin(2z)\sec^2z}{\tan x + \tan y + \tan z} = 2\left(\frac{ \tan x + \tan y + \tan z}{\tan x + \tan y + \tan z}\right) = 2$$

Which clearly equals the RHS.
We have that

$$u(x,y,z)=\ln(\tan x+\tan y+\tan z)$$

Then

$$\frac{\partial u}{\partial x}=\frac{\sec^2x}{\tan x+\tan y+\tan z}\\
\frac{\partial u}{\partial y}=\frac{\sec^2y}{\tan x+\tan y+\tan z}\\
\frac{\partial u}{\partial z}=\frac{\sec^2z}{\tan x+\tan y+\tan z}$$

Then

$$\sin(2x)\frac{\partial u}{\partial x}+\sin(2y)\frac{\partial u}{\partial y}+\sin(2z)\frac{\partial u}{\partial z}=\\
\frac{\sin(2x)\sec^2x+\sin(2y)\sec^2y+\sin(2z)\sec^2z}{\tan x+\tan y+\tan z}=\\
\frac{2\sin x\cos x\sec^2x+2\sin y\cos y\sec^2y+2\sin z\cos z\sec^2z}{\tan x+\tan y+\tan z}=\\
2\frac{\tan x+\tan y+\tan z}{\tan x+\tan y+\tan z}=2$$