How Do You Compute Killing Vectors for a Given Metric?

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I want to find all the killing vectors of the metric ##x²dx² + xdy²##. We could guess somethings by intuition and check it, but i decided to use the equation itself. Unfortunatelly, i realized that i am not sure how to manipulate the equation

$$L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} =^{?} 0$$

When i wrote this equation, i was not sure how the index should work here, so i guessed that we need to check it for all possible combinations of ab. ##(a,b) = [(0,0),(0,1),(1,0),(1,1)]##$. So, seeing this way, we need to find ##\vec X = (X^{0},X^{1})##? THe answer is something in partial derivatives. How would i got that just with the equation? I would appreciate if you give a tip of how to really start the computations, because as you can see, i am stuck right at the start.
 
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The Killing equation for a metric $g_{ab}$ is given by$$L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} = 0.$$In this case, you are trying to find the Killing vectors $\vec{X}=(X^0,X^1)$ for the metric $$g_{ab}=\begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}.$$ To solve for $\vec{X}$, begin by computing the partial derivatives of the metric components with respect to each coordinate:$$\partial_0 g_{00} = \partial_0 (x^2) = 2x,$$$$\partial_1 g_{00} = \partial_1 (x^2) = 0,$$$$\partial_0 g_{01} = \partial_0 (0)=0,$$$$\partial_1 g_{01} = \partial_1 (0) = 0,$$$$\partial_0 g_{11} = \partial_0 (xy^2) = y^2,$$$$\partial_1 g_{11} = \partial_1 (xy^2) = 2xy.$$Now, substitute these partial derivatives into the Killing equation and solve for $\vec{X}$:\begin{align*}L_\chi g_{ab} &= g_{ad}\partial_b X^d + g_{bd}\partial_a X^d + X^e \partial_e g_{ab} \\&= \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\begin{bmatrix}\partial_0 X^0 & \partial_1 X^0 \\ \partial_0 X^1 & \partial_1 X^1\end{bmatrix} + \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\begin{bmatrix}\partial_0 X^0 & \partial_1 X^0 \\ \partial
 
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