The question itself (Pressey, Elementary Differential Geometry, Exercise 6.1.4) has been posted before, but the previous question only dealt with the reparameterization portion and not the question itself.
Suppose that a surface patch $\tilde{\sigma}(\tilde u,\tilde v)$ is a reparametrization of a surface patch $\sigma (u,v)$.
Let $$\tilde E d\tilde u^2+ 2\tilde F d\tilde ud\tilde v+\tilde G d\tilde v^2$$ $$Edu^2+2Fdudv+Gdv^2$$ be their first fundamental form.
Show that $$du=\frac{\partial u}{\partial \tilde u}d\tilde u+\frac{\partial u}{\partial \tilde v}d\tilde v$$
$$dv=\frac{\partial v}{\partial \tilde u}d\tilde u+\frac{\partial v}{\partial \tilde v}d\tilde v$$
I have gotten as far to show that $\tilde{E}d\tilde u^2+2\tilde Fd\tilde u\tilde v+\tilde Gd\tilde v^2=E\left(\frac{\partial u}{\partial\tilde u}d\tilde u+\frac{\partial u}{\partial\tilde v}d\tilde v\right)^2+2F\left(\frac{\partial u}{\partial\tilde u}d\tilde u+\frac{\partial u}{\partial\tilde v}d\tilde v\right)\left(\frac{\partial v}{\partial\tilde u}d\tilde u+\frac{\partial v}{\partial\tilde v}d\tilde v\right)+G\left(\frac{\partial v}{\partial\tilde u}d\tilde u+\frac{\partial v}{\partial\tilde v}d\tilde v\right)^2$
I'm not sure where to go from here, and any help would be appreciated.
The fundamental form is a red herring; this is just about the algebraic relationship between forms and tangent vectors.
In particular, since $\frac{\partial}{\partial \bar{u}}$ and $\frac{\partial}{\partial \bar{v}}$ are a basis for the tangent space, a differential form is completely determined by its products with these two tangent vectors.
E.g. if we take the first equation and apply these differential forms:
$$ \frac{\partial}{\partial \bar{u}} \cdot \mathrm{d}u = \frac{\partial u}{\partial \bar{u}}$$ $$ \frac{\partial}{\partial \bar{u}} \cdot \left( \frac{\partial u}{\partial \bar{u}} \mathrm{d} \bar{u} + \frac{\partial u}{\partial \bar{u}} \mathrm{d} \bar{v} \right) = \frac{\partial u}{\partial \bar{u}} \frac{\partial \bar{u}}{\partial \bar{u}} + \frac{\partial u}{\partial \bar{u}} \frac{\partial \bar{v}}{\partial \bar{u}} = \frac{\partial u}{\partial \bar{u}} \cdot 1 + \frac{\partial u}{\partial \bar{u}} \cdot 0 $$ and so both sides of the equation reduce to the same scalar field. This is also true when applying $\frac{\partial}{\partial \bar{v}}$. Since both sides give the same products, they must be equal forms: $$ \mathrm{d}u = \frac{\partial u}{\partial \bar{u}} \mathrm{d} \bar{u} + \frac{\partial u}{\partial \bar{u}} \mathrm{d} \bar{v} $$
Note that it doesn't even matter that $(u,v)$ is a coordinate system; all that matters in the above calculation is that $u$ is a scalar field and that $(\bar{u}, \bar{v})$ is a coordinate system.