I was discussing with some friend about how to define the tangent bundle for a smooth variety since this is very natural in the manifold setting and we couldn't find references discussing in detail.
For example, if $X\subset\mathbb{A}^n$ is an affine algebraic variety we can define the tangent bundle $TX\subset\mathbb{A}^{2n}$, where first $n$ coordinates respect de equations for $X$ and the other the equations for tangent spaces: $\sum_{i=1}^n{\frac{\partial f}{\partial x_i}}(x)y_i=0$. These imply that $TX$ is an algebraic variety, but my problem is: is this a vector bundle? The problem is that I don't see how it will be locally trivial. Zariski open sets are very large and it will impose a lot of rigidity; also there is no implicit function theorem so we can get trivializations for $TX$ like in complex analytic viewpoint.
If this is locally trivial, how can it be shown? Otherwise, can an example be easily found?
Yes, what you ask for is true. For $X$ a smooth variety over $k$, $TX$ is the dual of the cotangent bundle, the locally free sheaf $\Omega^1_{X/k}$, and is thus again locally free. As locally free sheaves correspond to (locally trivial) vector bundles, this gives the result you're asking for.
The proof that the cotangent sheaf on a smooth variety is locally free is a classic and can be found in most algebraic geometry texts. Here's a summary of StacksProject's proof, which can be found here.
Any smooth variety $X/k$ can be covered by affine opens of the form $\operatorname{Spec} R$ with $R=k[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$, such that the determinant of the matrix $\left(\frac{\partial f_i}{\partial x_j}\right)$ for $1\leq i,j\leq m$ is a unit in $R$. Then $\Omega^1_{R/k}$ is free on the classes $df_i$, and this shows that $\Omega^1_{X/k}$ is locally free.