Firstly, let's invert this equation so we don't have to do implicit differentiation.

\(\displaystyle{R}={\frac{{{1}}}{{{\frac{{{2}}}{{{r}{1}}}}+{\frac{{{1}}}{{{3}{r}{2}}}}}}}={\frac{{{1}}}{{{\frac{{{6}{r}{2}+{r}{1}}}{{{3}{r}{1}{r}{2}}}}}}}={\frac{{{3}{r}{1}{r}{2}}}{{{6}{r}{2}+{r}{1}}}}\)

To see if this function is increasing in r1, we need to take its derivative with respect to r1 and check that it is nonnegative for every value of r1.

So let's do that. Remember to use the quotient rule.

\(\displaystyle{R}'{\left({r}{1}\right)}=\frac{{{\left({3}{r}{2}\right)}{\left({6}{r}{2}+{r}{1}\right)}-{3}{r}{1}{r}{2}{\left({1}\right)}}}{{\left({1}\right)}^{{2}}}={\left({18}{r}{2}^{{2}}\right)}+{3}{r}{2}{r}{1}-{3}{r}{2}{r}{1}={18}{r}{2}^{{2}}\)

Because r2>0, we see that R'(r1) > 0 everywhere. Hence R is an increasing function of r1 Q.E.D.

\(\displaystyle{R}={\frac{{{1}}}{{{\frac{{{2}}}{{{r}{1}}}}+{\frac{{{1}}}{{{3}{r}{2}}}}}}}={\frac{{{1}}}{{{\frac{{{6}{r}{2}+{r}{1}}}{{{3}{r}{1}{r}{2}}}}}}}={\frac{{{3}{r}{1}{r}{2}}}{{{6}{r}{2}+{r}{1}}}}\)

To see if this function is increasing in r1, we need to take its derivative with respect to r1 and check that it is nonnegative for every value of r1.

So let's do that. Remember to use the quotient rule.

\(\displaystyle{R}'{\left({r}{1}\right)}=\frac{{{\left({3}{r}{2}\right)}{\left({6}{r}{2}+{r}{1}\right)}-{3}{r}{1}{r}{2}{\left({1}\right)}}}{{\left({1}\right)}^{{2}}}={\left({18}{r}{2}^{{2}}\right)}+{3}{r}{2}{r}{1}-{3}{r}{2}{r}{1}={18}{r}{2}^{{2}}\)

Because r2>0, we see that R'(r1) > 0 everywhere. Hence R is an increasing function of r1 Q.E.D.