Is this calculation for the 'free' climbing speed when flying in a headwind with a gradient correct?
I propose to look at it in a different way. Let's assume a constant airspeed climb.  

We have two scenarios: one where there is no headwind gradient, and one with a headwind gradient. We can calculate the achieved climb speed in both scenarios and then compare them to see what the effect of the wind is. By selecting a constant airspeed, the drag is constant in both scenarios so they can be easily compared. 

In both scenarios the aircraft will have the same constant thrust, such that the thrust exceeds the drag. The excess thrust is equal to 5% of the weight.

For the purpose this answer, I do not look at calibrated airspeed but only at true airspeed. Basically I neglect effects of changing density with altitude.

Also, the angle of climb (flight path angle $\gamma$) is sufficiently small to be in the linear domain (such that $\cos\gamma \approx1$ and  $\sin\gamma \approx \gamma$)

Scenario 1: climbing in a constant wind field

In the first scenario the aircraft is climbing in a constant wind field. This can be no wind at all, or a constant headwind or a constant tailwind. 

For the aircraft to perform a constant airspeed climb in this constant wind field, all forces acting on the aircraft (thrust, drag, lift, weight) need to be in equilibrium.  




Thrust = green
Drag = red
Lift = blue
Weight = black


$\frac{dV}{dt} = \frac{T - D - W sin \gamma}{m} = g\left( \frac{T - D}{W}  - \sin \gamma\right) = 0 $

With a thrust excess ($T-D$) of 5% of the weight, the aircraft need to climb at a flight path angle of 0.05 radians which is approximately 2.86 degrees.

$\frac{T - D}{W} = \sin \gamma = 0.05$

The vertical speed $w$ is then equal to:

$w = V\cdot\sin\gamma $

So in the first scenario the aircraft will achieve a climb speed that is equal to 5% of the true airspeed. Taking 60 m/s from your calculation for the airspeed, the aircraft will climb at 3 m/s = 591 ft/min. 

Scenario 2: climbing in a wind gradient

In our second scenario, the same aircraft climbs with the same velocity and same thrust setting, but now it encounters a headwind gradient of 10 m/s per 100 meter of climb. This equates to a wind gradient of 0.1/s:

$\tau = \frac{10 \textrm{ m/s} }{100 \textrm {m}} = 0.1 \frac{1}{\textrm{s}}$.

A climb of 3 m/s would result in an effective wind increase of $\tau \cdot w$ = 0.3 m/s2 which is effectively an increase of true airspeed. Therefore, this isn't a constant speed climb anymore. 

To make sure the climb is performed at a constant air speed, the aircraft needs to climb faster.

$\frac{dV}{dt} = \frac{T - D - W sin \gamma}{m} = g\left( \frac{T - D}{W}  - \sin \gamma\right) + \tau V \sin \gamma = 0 $

$\sin \gamma =  \frac{g}{g-\tau V}\cdot\frac{T-D}{W}= \frac{9.81}{9.81-0.1\cdot 60}\cdot0.05 = 0.129$

$w=V\sin\gamma =$ 7.72  m/s = 1521  ft/min



In the wind gradient the climb speed of 1521 ft/min is considerably higher than the 591 ft/min achieved in a constant wind field with the same power setting. The difference is a factor of 2.57 which can be attributed to the wind.