APL+WIN, 5 bytes.

Prompts for input.

÷⎕*.5

Try it online! Thanks to Dyalog Classic

Python 3, 30 bytes

Self answer ;) (and actually my first!)

Solution: (could defo be golfed futher) print(1/(float(input()))**0.5)

Breakdown:

print() Self explanatory

1/() Divide 1 by the square root, with brackets to ensure the order works (could defo be golfed)

float(input()) Get input, convert to float

**0.5 Apply the power of 0.5 to the inputted float, which is equivalent to a square root, but doesn't require importing the maths library.

Maple, 22 bytes

x->fsolve(x*u^2-1)[2];

Not sure if this is cheating, but it doesn't involve builtin square roots or the like. Older versions of Maple only returned one real root, but now all real roots of a polynomial are returned, so we need the [2] to select the positive root.

Vyxal 3, 2 bytes

√⅟

Vyxal It Online!

boring answer. I tried implementing the actual Q_rsqrt but vyxal has no support for binary conversion of floats, required for the "evil floating point hack".

<script type="vyxal3">
√⅟
</script>
<script>
    args=[["0.15625"]]
</script>
<script src="https://themoonisacheese.github.io/snippeterpreter/snippet.js" type="module"/>

JavaScript (Node.js), 35 bytes

x=>(g=y=>y*y*x<1?y:g(y*.993))(9e99)

Try it online!

Need stack extend