Let $S \subseteq \mathbb{R}^3$ be a surface and $p \in S$. Given any basis $\{v_1,v_2\}$ of the tangent plane, there exists some parameterization $\varphi: U \to S$ centered at $p$ such that $\frac{\partial \varphi}{\partial u}=v_1$ and $\frac{\partial \varphi}{\partial v}=v_2$?
In general, if I take $v_1$, there exists a curve $\alpha: (-\epsilon,\epsilon) \to S$ such that $\alpha(0)=p$ and $\alpha'(0)=v_1$. Similarly for $v_2$. If $\psi$ is a given parameterization whose image contains the trace of the curve $\alpha$, I consider $d\psi^{-1}(v_1)=w_1$ that clearly belongs to $T_p(U)=\mathbb{R}^2$. Then $w_1$ may be written as linear combination of the basis of $T_q(U)$, that we denote by $\frac{\partial }{\partial x},\frac{\partial}{\partial y}$. Repeating the argument for $v_2$, we obtain a matrix $A$ whose rows represent the coordinates of $w_1$ and $w_2$ with respect to the basis of $T_q(U)$. Is the map $\varphi:=\psi \circ A$ the desired parameterization? May you express the direct computations for $\frac{\partial (\psi \circ A)}{\partial u}$, please?