How to estimate Reliability function in Weibull by the failure rate

Question

The Weibull failure rate function, is given by:

$$\lambda(T) = \frac{f(T)}{R(t)}=\frac{\beta}{\eta}\left(\frac{T-\gamma}{\eta}\right)^{\beta-1}$$

If I have $f(T)$ values and $\lambda(T)$ the failure rate values.

can I calculate $R(t)$ as?

$$R(t) = \frac{f(T)}{\lambda(T)}$$

More details:

My items follow Weibull and in my research design I would like to find a way to estimate the reliability of these items by using the failure rate, if the formula above is statistically right.

One way to estimate the reliability by

$$R(t) = 1- {f(T)}$$

But I am not interested in using this equation, since failure rate can not be used

My experimental design divided into two steps.

1- Find the failure rate of machines (item).

2- Using this failure rate to estimate the reliability of these machines (item).

I will have no issues if machines follow exponential distribution, because failure rate is the parameter of reliability function, however in machines that follow Weibull, failure rate is not in reliability function only shape and scale can be used to estimate reliability

Example of failure data of one physical machines in hours and the order of failure. For the simplicity let's assume this machine follow Weibull

failure_time Order 
    240.3   1
    244.9   2
    247.6   3
    256.5   4
    277.6   5
    278.3   6
    280.7   7
    281.8   8
    343.4   9
    349.6   10
    408.4   11
    409.6   12
    419.3   13
    425.8   14
    438.3   15
    464.4   16
    464.7   17
    516.1   18
    521.5   19
    522.4   20
    650.0   21
    650.9   22
    653.7   23
    656.0   24
    662.4   25
    664.4   26

I apply non parametric method to estimate the parameters, and I have

Scale = 488.9612 Shape = 3.136 

by theses parameter I can find $f(t)$, and I already have the Instantaneous Failure Rate $\lambda(T)$ from first step.

Now I can use these giving Equation to estimate the reliability

$$R(t) = \frac{f(T)}{\lambda(T)}$$

I am not sure if this could be right!