I have troubles with the following:
Consider the Lie Algebra $\text{sl}(2,\mathbb{R})$ of $\text{SL}(2,\mathbb{R})$. Let $$a_x: \text{sl}(2,\mathbb{R})\to \text{sl}(2,\mathbb{R}):Y\mapsto [X,Y].$$ The Killing form is $\langle x,y\rangle_k=\mathrm{tr}(a_x a_y)$. Show that $(\mathrm{sl}(2,\mathbb{R}), \langle\,,\,\rangle_k)$ is isomorphic to $(\mathbb{R}^3,\langle\, , \, \rangle_y)$, where $\langle(x,y,z),(u,v,t)\rangle_y=xu+yv-zt.$
I couldn't find an isomorphism between the Lie algebras.
This is just an explicit computation. Pick a basis $ A, B, H $ for the Lie algebra $ \mathfrak{sl}_2(\mathbf R) $ such that
$$ [A, B] = H, [H, A] = -2A, [H, B] = 2B $$
The adjoint map $ \textrm{ad}_X: Y \to [X, Y] $ of $ X = aA + bB + cH $ then has an explicit representation
$$ \begin{bmatrix} -2c & 0 & 2a \\ 0 & 2c & -2b \\ -b & a & 0 \end{bmatrix} $$
Now let $ X = a_1 A + b_1 B + c_1 H, \, Y = a_2 A + b_2 B + c_2 H $. Explicitly computing the trace of the matrix product $ \textrm{ad}_X \circ \textrm{ad}_Y $ gives
$$ \operatorname{tr}(\textrm{ad}_X \circ \textrm{ad}_Y) = 4 c_1 c_2 - 2 a_1 b_2 + 4 c_1 c_2 - 2 b_1 a_2 - 2 b_1 a_2 - 2 a_1 b_2 = 8 c_1 c_2 - 4 a_1 b_2 - 4 b_1 a_2 $$
which can be written, under a change of coordinates, in the form
$$ 8 c_1 c_2 - 2 (a_1 + b_1) (a_2 + b_2) + 2 (a_1 - b_1) (a_2 - b_2) $$
Now just define the isomorphism by $ aA + bB + cH \to \sqrt{2} \cdot (2c, a-b, a+b) $.