Is it possible to compute the y-coordinate of a point on SECP256K1, given only the x-coordinate

Given an $x$-coordinate of a point on the SECP256K1 curve, is it possible to calculate the corresponding $y$-coordinate?

Yes, if there exists such $y$ for the given $x$. And, absent other indication, such $y$ can only be found within sign (or equivalently, parity). That limitation is because if $y^2equiv x^3+7pmod p$ with $p=2^{256}−2^{32}−2^{10}+2^6-2^4−1$ as in secp256k1 has a solution $y_0$ in $[0,p)$, then $y_1=p-y_0$ also is a solution.

Note: in some cases including secp256k1 as used in Bitcoin, a public key with $x$ and without $y$ (that is, in compressed form) comes with a prefix of 02 if $y$ is even, 03 if $y$ is odd, and that allows to fully recover $y$.

By Euler's criterion, $x^3+7$ has a square root modulo $p$ if and only if $(x^3+7)^{(p-1)/2}bmod p=1$. That holds for the question's $x$, thus there are solutions.

In the general case, the Tonelli–Shanks algorithm can be used to find modular square roots. Since $pequiv3pmod4$, that algorithm reduces to computing $y_0gets (x^3+7)^{(p+1)/4}bmod p$ and $y_1gets p-y_0$. The question's $y$ happens to be $y_1$.

Justification: when we have checked $(x^3+7)^{(p-1)/2}bmod p=1$, and computed $y_0$ as $(x^3+7)^{(p+1)/4}bmod p$, the later is such that $$begin{array}{} {y_0}^2 &equiv&left((x^3+7)^{(p+1)/4}right)^2 &pmod p \ & equiv&(x^3+7)^{(p+1)/2} &pmod p \ &equiv&(x^3+7)^{(p-1)/2},(x^3+7)&pmod p\ &equiv&x^3+7 &pmod p & text{since};(x^3+7)^{(p-1)/2}bmod p=1end{array}$$ thus $y_0$ is a solution to $y^2equiv x^3+7pmod p$.

Definitions: $$begin{array}{l} uequiv vpmod p&underset{text{def}}iff v-u;text{ is a multiple of };p\ u=vbmod p&underset{text{def}}iff v-u;text{ is a multiple of };p;text{ and }0le u<p; end{array}$$