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In every book I have, these are solved like “we guessed this solution”, but I’m sure there is analytic way to solve them:

$y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y} = 0$

$(x + 2y)\frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} = 0$

There are more similar equations but I guess I just need to see at least one analytically solved.

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    method of characteristics https://en.wikipedia.org/wiki/Method_of_characteristics – 2017-01-11

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This is called a linear first-order homogeneous PDE. It is usually solved with characteristic method as follows:
0. Let equation $a_1(x, y)\frac{\partial{u}}{\partial{x}}+a_2(x,y)\frac{\partial{u}}{\partial{x}} $ be given.
1. Set $ x = x(t), y = y(t) $
2. Set what stands before $\frac{\partial{u}}{\partial{x}}, a_1$ in our case, to be equal $x_t$ and what stands before $\frac{\partial{u}}{\partial{y}}, a_2$ in our case, to be equal $y_t$
3. Use the chain rule to obtain: $\frac{du}{dt} = \frac{\partial{u}}{\partial{x}}x_t+\frac{\partial{u}}{\partial{y}}y_t$
4. Original equation equals zero if and only if $\frac{du}{dt} =0 $, or because of property of the full differential if $ x_t = a_1(x,y) $ and $y_t = a_2(x,y)$.
5. Solve the system of ODE's
6. Profit!

Append:

Let's solve your first PDE:
0. $y\frac{\partial{u}}{\partial{x}} - x\frac{\partial{u}}{\partial{y}}=0$
1. $x=x(t), y=y(t) $
6. Through steps 2-5 we otain the system of ODE's: $ x_t=y, y_t=-x $
Let's convert this system of ODE's to single ODE of second order (I think you're familiar how to solve such):
1. Differentiate the first eqution and substite it into the second one.
2. Obtain: $x_{tt}+x=0$
3. Solution to this equation is $c_1sin(t)+c_2cos(t)$
4. The y is: $y(t) = x_t = c_1cos(t)-c_2sin(t)$

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    So, basically I will get $x = yt + c_1$ and $y = xt + c_2$, so $u(x, y) = y^2 - c_2y + c_1x - x^2$, but that's not solution for any $c_1, c_2$, what do I do about it? edit: I'm talking about first equation – 2017-01-11
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    @PanMiroslav you are forgetting that y is also a function of t – 2017-01-11
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    oh well, so I get $x(t) = \int y(t)dt$ and $y(t) = -\int x(t)dt$, so what's next? I know that sinus and cosinus satisfy this, but how do I find it out? – 2017-01-11
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    yes sin and cosinus. How do you find it out? If you wish to be simplistic, just prove by substitution and use superposition priciple. Alternatively, use theory of ordinary linear differential equations. Their solutions are always of the form $e^{wt}$, where $w$ may be complex. – 2017-01-11
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They describe directional derivatives in the direction of the vector $\vec{V}=(a,b)$ if the general PDE is of the form $$ a(x,y) u_x + b(x,y) u_y = 0.$$ NOTE: if constant coefficients the general solution is of the form $f(bx - ay)$. This is known as the geometric method. However, we have variable coefficients.

With respect to the first PDE in your question The equations states that the characteristic curves are in the direction $(y,-x)$ with slopes of the characteristic curves given by $dy/dx = -x/y$ which can be solved for $y=\sqrt{C-x^2}$. The PDE says that the solution $u(x,y)$ is constant along the curves given by $y$ so that $u(x,y)=f(C)$. Find $C$ and this is the general solution.

This method works particularly well for PDEs of the form $$ \nabla u \cdot (x,y) = 0. $$ Also note that the geometric method works great in this case too.