How to control output resistance with voltage

You could do something like this:

^{simulate this circuit – Schematic created using CircuitLab}

The FF is to give a 50% duty cycle output (at half the frequency).

But you can make a good VCO without the 555 at all.

you only really need to change the current thart flows through R1s location, so there's many options. here's four of them

^{simulate this circuit – Schematic created using CircuitLab}

They'll akll give slightly different performance and with the exception of the second they all give the effect of lower resistance with lower voltage on "in". option 4 will give the best repeatability.

There are digital potentiometer ICs available that you could use in place of R1. You have to communicate with them using a micro-controller or other device however. Here is one example:

https://www.mouser.com/ProductDetail/Microchip-Technology/MCP4132-103E-P?qs=hH%252BOa0VZEiBueXVavvhYaw%3D%3D&gclid=EAIaIQobChMInJrjj7jh6gIVjJ-zCh1IrQoVEAQYASABEgLkdfD_BwE

It sounds like you are creating a Voltage Controlled Oscillator (VCO) with a 555.

Here is a guide for just such a project..

You cannot change the resistance of a resistor. Even in AC resistance of a resistor is independent of the frequency of AC.

However what you could do is use a BJT transistor. By applying a varying voltage at the base the collector current changes.

^{simulate this circuit – Schematic created using CircuitLab}

With increase in voltage of A , more current will flow through the load and we can say by Ohms law , since the ratio V/I has changed( assuming W is at fixed voltage) resistance has been changed ,more precisely it has been decreased.

The point of the arrow is to the ground the transistor is falsely oriented and because my computer is being serviced I just add this note.

If VA >0.65V then to find current in Rb we will write down Ohms law with a voltage drop of 0.65V:Ib=VA-0.65/Rb.

The current through Rl will be k times the current of the base where k is the gain of the BJT and it is unique for every transistor.