If $w$ is smooth and compactly supported in $[1/2,2]$, say, then $$\sum_{n}\frac{\varphi(n)}{n}\frac{\varphi(n+1)}{n+1}w(n/x) = Cx + O(1).$$ Write $$\frac{\varphi(n)}{n} = \sum_{d\mid n} \frac{\mu(d)}{d}$$ and $$\frac{\varphi(n+1)}{n+1} = \sum_{e \mid n+1} \frac{\mu(e)}{e},$$ then interchange the orders of summation. Perform Poisson summation on the $n$ variable to turn the sum into something like $$\sum_d \frac{\mu(d)}{d} \sum_{(e,d)=1}\frac{\mu(e)}{e} \frac{x}{d^2e^2}\sum_{|k|\leq de/x} e\left(-k\frac{\overline{d}}{e}\right)\widehat{w}(kx/de),$$ where $\overline{d}$ denotes the inverse of $d$ modulo $e$. The $k=0$ term gives the main term, which is seen to be $Cx + O(1)$.
The nonzero frequencies contribute an error of size $O_A((\log x)^{-A})$. For these terms, break $d$ and $e$ into dyadic ranges $d \asymp D, e \asymp E$. Clearly we may assume $DE \gg x$. Also, we may assume without loss of generality that $D \leq E$, otherwise use the reciprocity relation $$\frac{\overline{d}}{e} + \frac{\overline{e}}{d}\equiv \frac{1}{de}\pmod{1}$$ to switch $d$ and $e$ in the exponential. After some work, Siegel-Walfisz and the large sieve cover the case when $D\leq E (\log x)^{-B}$, so we may assume $D \approx E$. The double sum over $d$ and $e$ may then be suitably bounded using results of Duke-Friedlander-Iwaniec on bilinear forms in Kloosterman fractions (see also work of Bettin and Chandee).