I am studying about Killing vector from an online source as follows:
I couldn't obtain the six equations (50 - 55) from equation (49). Also, I couldn't find $\partial^2_{x} \xi^{y}=0$ as mentioned (by applying $\partial_{x}$ to equation (52) and use (51)). May someone show the detail steps to solve the six Killing vectors. Thank you so much.
In the usual three dimensional Euclidean space in Cartesian coordinates, one has $g_{\mu \sigma} = \delta_{\mu \sigma}$. This is a constant, so its partial derivatives are zero. This leads to the equalities $g_{\mu \sigma} \partial_{\rho}\xi^{\mu} = \partial_{\rho}\xi^{\sigma}$, $g_{\rho \nu}\partial_{\sigma} \xi^{\nu}= \partial_{\sigma}\xi^{\rho}$, and $\partial_{\mu}g_{\rho \sigma}= 0$. So that (49) becomes, in this specific context $$ \forall \rho, \sigma \in \{x,y,z\}, \quad \partial_{\rho}\xi^{\sigma} + \partial_{\sigma}\xi^{\rho} = 0. $$ Setting $\rho = \sigma = x$ gives (twice) $(50)$. Setting $\rho = x$, $\sigma = y$ gives $(51)$. Setting $\rho = x$, $\sigma = z$ gives $(52)$. Setting $\rho = \sigma = y$ gives (twice) $(53)$. Setting $\rho = z$, $\sigma = y$ gives $(54)$. Finally, setting $\rho= \sigma = z$ gives (twice) $(55)$.
Regarding the last claim: there is a typo in the document. You should use $(50)$ (and not $(51)$) to obtain $\partial_x^2 \xi^y = 0$, after having applied $\partial_x$ to $(51)$ (and not $(52)$). Indeed, partial derivatives commute, so that $\partial_x (\partial_y \xi^x) = \partial_y (\partial_x \xi^x) = 0$ by $(50)$.
Edit. Let $\xi$ be a solution. Then as we have seen, we have $$ \partial_x\xi^x = \partial_y\xi^y = \partial_z\xi^z = 0 $$ and $$ \partial^2_x\xi^y = \partial^2_x \xi^z = \partial^2_y\xi^x = \partial^2_y\xi^z = \partial^2_z\xi^x = \partial^2_z\xi^y =0. $$ This means that:
Thus, there are nine constants $a^x,b^x,c^x,a^y,b^y,c^y,a^z,b^z$ and $c^z$ such that \begin{align} \xi^x &= a^x\phantom{x} + b^x y + c^x z, \\ \xi^y &= a^y x + b^y \phantom{y} + c^y z, \\ \xi^z &= a^z x + b^z y + c^z \phantom{z}. \end{align} But equations $(51)$, $(52)$ and $(54)$ now give $$ a^y = -b^x, \quad a^z = -c^x, \quad \text{and} \quad b^z = -c^y. $$ Therefore, there exist six constants $c_1,c_2,\ldots,c_6$ such that $$ \begin{align} \xi^x &= \phantom{-} c_1\phantom{x} -c_4 y + c_6 z, \\ \xi^y &= \phantom{-} c_4 x + c_2 \phantom{y} - c_5 z, \\ \xi^z &= -c_6 x + c_5 y + c_3 \phantom{z}. \end{align} $$ The set of Killing vector fields is thus contained in the six dimensional space spanned by the vector fields $\{\xi_1,\ldots,\xi_6\}$, where $\xi_j$ is chosen such that $c_j = 1$ and $c_k = 0$ if $k\neq j$. You can check that indeed, these vector fields are Killing, so that we have found all of them.
This is geometrically clear what these vector fields represent: $\xi^1,\xi^2$ and $\xi^3$ generate constant speed translations along the $x$-axis, $y$-axis and $z$-axis respectively, while $\xi^4$, $\xi^5$ and $\xi^6$ generate constant speed rotations about the $z$-axis, $x$-axis and $y$-axis respectively.
Note that this was expectable: we know that the isometries of the affine space $\Bbb R^3$ are precisely the rotations-translations. So we could have guessed in the first place that these six linearly-independent vector fields are Killing by just checking that they satisfy the Killing field equation. Since the space of Killing vector fields is of dimension at most $6(=\frac{3(3+1)}{2})$, this would have completed the proof, without solving any PDE!