So I know that this involves using the chain rule, but is the following attempt at a proof correct.
Let $M$ be an $n$-dimensional manifold and let $(U,\phi)$ and $(V,\psi)$ be two overlapping coordinate charts (i.e. $U\cap V\neq\emptyset$), with $U,V\subset M$, covering a neighbourhood of $p\in M$, such that $p\in U\cap V$. Consider a function $f:M\rightarrow\mathbb{R}$, and let $x=\phi(p)$, $y=\psi(p)$. It follows then that $$\frac{\partial f}{\partial x^{\mu}}(p)=\frac{\partial}{\partial x^{\mu}}\left((f\circ\phi^{-1})(\phi(p))\right)=\frac{\partial}{\partial x^{\mu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\\ \qquad \quad\; =\frac{\partial}{\partial y^{\nu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\frac{\partial}{\partial x^{\mu}}\left[\left((\psi\circ\phi^{-1})^{\nu}(\phi(p))\right)\right]\\ =\frac{\partial f}{\partial y^{\nu}}(p)\frac{\partial y^{\nu}}{\partial x^{\mu}}(p)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,$$ where $y(x)=(\psi\circ\phi^{-1})(\phi(p))$.
Hence, as $f$ is an arbitrary differentiable function, we conclude that $$\frac{\partial }{\partial x^{\mu}}=\frac{\partial y^{\nu}}{\partial x^{\mu}}\frac{\partial }{\partial y^{\nu}}$$ From this, we note that as $\lbrace\frac{\partial }{\partial x^{\mu}}\rbrace$ and $\lbrace\frac{\partial }{\partial y^{\nu}}\rbrace$ are two coordinate bases for the tangent space $T_{p}M$ at the point $p$, the two bases must be related by the formula above.