Aircraft performance charts all work the same way. Unfortunately, this one only gives the climb speed at 85 knots indicated airspeed (KIAS).
To find the climb speed at different speeds needs more knowledge about the airframe. Then you can apply a simple approximation following this procedure.
The chart does not tell where on the polar the aircraft is. Will it climb better when flying faster or not? This is impossible to tell. Since you have only one speed and one power setting, but a range of masses, the aircraft will not be at its optimum climb setting for most points. We can make an assumption and declare the reference mass of 1700 kg as the point where the cited conditions are at the optimum. But then I would need at least the zero-lift drag and the aspect ratio to make further assumptions.
From the previous answer we take the climb speed equation $$v_z = \frac{v}{C}\cdot sin\gamma = \frac{v}{C}\cdot\frac{T-D}{m\cdot g}$$
and set the correction factor C = 1 for now. The resulting error is small at low speeds. Now we need drag and thrust.
First drag: The drag coefficient $c_D$ is approximately
$$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
with the known aspect ratio $AR$ and an assumed Oswald factor $\epsilon$ of 0.85. To arrive from here at the drag we need to multiply this with the dynamic pressure $q = ½\rho\cdot v^2$ and the reference area $S$:
$$D = ½\rho\cdot v^2\cdot S\cdot c_{D0} + \frac{(1700\cdot g)^2}{½\rho\cdot v^2\cdot S\cdot\pi\cdot AR\cdot\epsilon} = ½\rho\cdot v^2\cdot S\cdot c_{D0} + \frac{(1700\cdot g)^2}{½\rho\cdot v^2\cdot\pi\cdot b^2\cdot\epsilon}$$
when we substitute $AR = b^2/S$ with $b$ the span of your airplane. As I said before, induced drag depends on span, not on aspect ratio.
Now for the thrust. In a propeller aircraft power is constant and thrust is inverse with airspeed. Not indicated, but true airspeed, so we need to be careful. At sea level both are equal, and then thrust is straightforward: $$T = \left(½\rho\cdot v_{ref}^2\cdot S\cdot c_{D0} + \frac{(1700\cdot g)^2}{½\rho\cdot v_{ref}^2\cdot\pi\cdot b^2\cdot\epsilon} + \frac{5.5}{v_{ref}}\cdot1700\cdot g\right)\cdot\frac{v_{ref}}{v}$$
with $v_{ref}$ = 43.7278 m/s, which is 85 kts in sane units. The first two terms in the bracket look familiar: They are the drag contribution. The third term accounts for the climb speed of 5.5 m/s at the reference point in sea level altitude, so this accounts for the change in potential energy. If you need the thrust at higher altitudes, please correct $v_{ref}$ with the square root of the density ratio.
If those equations look daunting, continue solving for the climb speed:
$$v_z = \frac{v}{C}\cdot\frac{\left(D + \frac{v_{z_{ref}}}{v_{ref}}\cdot m\cdot g\right)\cdot\frac{v_{ref}}{v} - D}{m\cdot g}$$
$$v_z = \frac{1}{C}\cdot\left(\frac{D\cdot\left(v_{ref}-v\right)}{m\cdot g} + v_{z_{ref}}\right)$$
