Please be patient as I struggle with latex here .....

Part 1 of the problem says to start with:

$ \frac{\partial\bar{r}}{\partial{q}_{1}} ={h}_{1} \hat{q}_{1} $ and then to find an expression for $ {h}_{1} $ that agrees with $ {g}_{ij}=\sum_{l} \frac{\partial{x}_{l}}{\partial{q}_{i}}\frac{\partial{x}_{l}}{\partial{q}_{j}} $

My attempt is:

$ \hat{q}_{1} = \frac{1}{h}_{1} \frac{\partial\bar{r}}{\partial{q}_{1}}$

but $ \hat{q}_{1}.\hat{q}_{1}=1$

Then $ \left({h}_{1}\right)^{\!{2}} =\left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)^{\!{2}}$

Now $ \left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)= \left(\frac{\partial{x}}{\partial{q}_{1}}\right)+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)+\left(\frac{\partial{z}}{\partial{q}_{1}}\right) $

so $ \left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)^{\!{2}}= \left(\frac{\partial{x}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{z}}{\partial{q}_{1}}\right)^{\!{2}} $

Now $ {g}_{ii} $ is defined = $ {h}_{i}^{\!{2}}$ and comparing with $ {g}_{ij}=\sum_{l} \frac{\partial{x}_{l}}{\partial{q}_{i}}\frac{\partial{x}_{l}}{\partial{q}_{j}} $ above...

Therefore $ {h}_{1}= \sqrt{ \left(\frac{\partial{x}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{z}}{\partial{q}_{1}}\right)^{\!{2}}} $ QED - but have I done anything illegal here?

-----

Part 2 is a derivation:

Again starting with $ \hat{q}_{1} = \frac{1}{h}_{1} \frac{\partial\bar{r}}{\partial{q}_{1}}$

Then $ \frac{\partial\hat{q}_{1}}{\partial{q}_{2}}=\frac{1}{{h}_{1}}\frac{\partial{}^{2}\bar{r}}{\partial{q}_{1}\partial{q}_{2}} = \frac{1}{{h}_{1}}\frac{\partial}{\partial{q}_{1}}\left(\frac{\partial\bar{r}}{\partial{q}_{2}}\right)^{\!{}} $ (${h}_{1}$ constant w.r.t. 2)

but $ \frac{\partial\bar{r}}{\partial{q}_{2}}={h}_{2} \hat{q}_{2} $

so $ \frac{\partial\hat{q}_{1}}{\partial{q}_{2}}=\frac{1}{{h}_{1}} \frac{\partial\left({h}_{2}\hat{q}_{2}\right)}{\partial{q}_{1}}= \hat{q}_{2}\frac{1}{{h}_{1}}\frac{\partial{h}_{2}}{\partial{q}_{1}}$

Which is QED - but again I have this uncertain feeling so would appreciate confirmation there I have done nothing untoward

-----------------

Part 3 has me so far, I would appreciate a hint ....

Derive $ \frac{\partial\hat{q}_{1}}{\partial{q}_{1}}= -\sum_{j\ne{1}}^{} \hat{q}_{2}\frac{1}{{h}_{2}}\frac{\partial{h}_{1}}{\partial{q}_{2}}$

I have tried a few things without success ... probably there is a trick I haven't encountered before?