It looks like i found an answer in book
Chandrasekhar S. The Mathematical Theory of Black. Vol. 1. Cambridge: Oxford Univ. Press, 1983. 107 p.
, but i still appreciate any help for answering some questions (that are well known for physicist) across the proof.
Let us have Schwarzschild metric
$$\mathrm{d} s^2=(1-2 M / r)(\mathrm{d} t)^2-\frac{(\mathrm{d} r)^2}{1-2 M / r}-r^2\left[(\mathrm{~d} \theta)^2+(\mathrm{d} \varphi)^2 \sin ^2 \theta\right]$$
Lagrangian in our case looks like this
$$\mathscr{L}=1 / 2\left[(1-2 M / r) \dot{t}^2-\dot{r}^2 /(1-2 M / r)-r^2 \dot{\theta}^2-\left(r^2 \sin ^2 \theta\right) \dot{\varphi}^2\right],$$ where the dot means differentiation by $\tau$.
We are interested in canonical impulse $p_t$:
$$
p_t=\frac{\partial \mathscr{L}}{\partial t}=\left(1-\frac{2 M}{r}\right)\dot{t} ,
$$
$$
\frac{\mathrm{d} p_t}{\mathrm{~d} \tau}=\frac{\partial \mathscr{L}}{\partial t}=0 \text{ (by definition?)},
$$
$$
p_t=\left(1-\frac{2 M}{r}\right) \frac{\mathrm{d} t}{\mathrm{~d} \tau}=\mathrm{const}=E,
$$
At this point everything became clear. In the problem there is no spherical movement, so Schwarzschild metric reduces to
$$\mathrm{d} s^2=(1-2 M / r)(\mathrm{d} t)^2-\frac{(\mathrm{d} r)^2}{1-2 M / r}$$
Since $ds^2 \text{(why?)}=c^2d\tau^2=\tau^2$ , so we have:
$$\mathrm{d} \tau^2=(1-2 M / r)(\mathrm{d} t)^2-\frac{(\mathrm{d} r)^2}{1-2 M / r}$$
And substituting $\frac{dt}{d\tau}$ into the expression we simply get:
$$
\left(\frac{d r}{d \tau}\right)^2=2 M / r-\left(1-E^2\right)
$$
I am not sure about all sign's here, but is it mostly right?
