Verifying a continued fraction related to $\logφ$. The continued fraction is the following,
$${1+\cfrac{1\cdot 2}{3φ+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7φ+\cfrac{3\cdot 4}{9+\ddots}}}}}=\frac{2}{3\logφ}\tag{1}$$
Where,
$$φ=\frac{1+\sqrt{5}}{2}$$
Something with I found in an backside of my previous notes where I kept my recreational math works. So can the closed form be verified in any sort (most preferably by using already established identities)?

This can be deduced from Gauss's continued fraction for $_2F_1$, written as $$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}.$$ With $a=b=1$ and $c=3/2$, computing the values of $_2F_1$, we obtain $$1+\cfrac{1\cdot 2}{3/z+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7/z+\cfrac{3\cdot 4}{9+\ddots}}}}=1+\frac{2z}{3}\frac{_2F_1(2,1;5/2;-z)}{_2F_1(1,1;3/2;-z)}=\frac{\sqrt{z(1+z)}}{\sinh^{-1}\sqrt{z}}.$$ Now we put $z=1/\varphi=\varphi-1$ and note that $$\sinh^{-1}\frac{1}{\sqrt\varphi}=\log\left(\frac{1}{\sqrt\varphi}+\sqrt{\frac{1}{\varphi}+1}\right)=\log\left[\sqrt\varphi\left(\frac{1}{\varphi}+1\right)\right]=\frac32\log\varphi.$$
