Let $\mathfrak{g}$ be a Lie algebra and let $[-,-]:\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ be the Lie bracket. By a (Lie algebra) deformation of $\mathfrak{g}$ one usually means a (typically, analytic) one-parameter family of Lie brackets $[-,-]_t$ on the same underlying vector space $\mathfrak{g}$, agreeing with the original Lie bracket at $t=0$. A deformation is trivial if $[-,-]_t$ is obtained from $[-,-]$ by a one-parameter group $g(t)$ of linear transformations of $\mathfrak{g}$. The derivative at $t=0$ of $[-,-]_t$ defines a skew-symmetric bilinear map $\phi: \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ which is a cocycle in the Chevalley-Eilenberg complex $C^2(\mathfrak{g};\mathfrak{g})$, whereas if the deformation is trivial, this cocycle is actually the coboundary of some cochain in $C^1(\mathfrak{g};\mathfrak{g})$. Therefore, infinitesimal nontrivial deformations are classified by the Chevalley-Eilenberg cohomology group $H^2(\mathfrak{g};\mathfrak{g})$. You can work out easily what the cocycle condition by taking the derivative at $t=0$ of the Jacobi identity for the deformed bracket $[-,-]_t$.
You can work in a basis $X_a$ for $\mathfrak{g}$ and then $[-,-]_t$ is defined by structure "constants" (which are now a function of $t$!): $C_{ab}{}^c(t)$ and the cocycle is the derivative at $t=0$ of these structure constants.
Given a nontrivial infinitesimal deformation, one is typically interested in whether it will integrate to a one-parameter family of deformations. There is an infinite number of obstructions to integrability: each defined provided the previous one is overcome, which are classes in $H^3(\mathfrak{g};\mathfrak{g})$.
So, for example, if you can calculate $H^2(\mathfrak{g};\mathfrak{g})$ and $H^3(\mathfrak{g};\mathfrak{g})$ you are part of the way to understanding the deformation theory of $\mathfrak{g}$. Computing Chevalley-Eilenberg cohomology is simply a question of solving linear equations, so it is doable in principle and particularly amenable to computer calculations. But there are several results of a general nature which can save you lots of computation. For example, if $\mathfrak{g}$ is semisimple, then $H^2(\mathfrak{g};\mathfrak{g}) = 0$, so that such algebras are (infinitesimally) rigid. This means that any deformation is really just a $t$-dependent change of basis.
In your comment you mention rotational symmetry. The rotation algebra in dimension $d>2$ is semisimple, so in fact it is rigid. Any infinitesimal deformation you find is certainly trivial.
You can read about this in the original paper: C. Chevalley, S. Eilenberg, Cohomology theory of Lie groups and Lie algebras, Trans. Amer. Math. Soc. 63, (1948). 85–124. It is very well written and not very abstract at all. You may also benefit from the references in the answers to this MathOverflow question.
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