I was revising chain rule and I made up a problem to write down in my notes that uses it at least two times. Here it is, if a function $\zeta(x) = (z(x))^2$ where $z(x) = x + f(x), f(x) = \ln(g(x))$ and $g(x) = \frac{1}{2}x^2$ then $\zeta'$ or $\frac{d\zeta}{dx}$ is defined as, \begin{align*} \zeta'(x) & = \frac{d\zeta}{dz}\times \frac{dz}{df}\times \frac{df}{dg} \times \frac{dg}{dx}\\ \zeta'(x) & = 2(z(x))z'(x) \\ & = 2(z(x))(1 + f'(x)) \\ & = 2(z(x))(1 + (\ln(g(x)))') \\ & = 2(z(x))\Big(1 + \Big(\frac{1}{g(x)}\Big)g'(x)\Big) \\ & = 2(z(x))\Big(1 + \Big(\frac{1}{g(x)}\Big)x\Big) \\ \end{align*}
Did I get it right?
Have a little more confidence! Your work is completely correct.
About writing it in Leibniz notation: we have $\zeta(x) = \varphi(y)$, where $\varphi$ is defined by $x \mapsto x^2$ and $h$ by $y \mapsto h(y) = y + f(y)$. Therefore: $$ \begin{aligned} \frac{\mathrm{d}\zeta}{\mathrm{d}x} &= \frac{\mathrm{d} \varphi}{\mathrm{d}y} \frac{\mathrm{d}y}{\mathrm{d}x} \\ &= \frac{\mathrm{d} \varphi}{\mathrm{d}y} \left(1 + \frac{\mathrm{d}f}{\mathrm{d}y} \right) \end{aligned}$$
And from here on you can use the definitions of $f$ and $g$ to expand the last equality in terms of $f$ and $g$, but I'd rather leave it at just the first one.