$\def\P{\mathbb P}$So let's first find the index used in defining $X$, that is define $ T(\omega) := \min \left\{k\in \mathbb N \mid U_k(\omega) \le f\bigl(V_k(\omega)\bigr)/60\right\} $ if this minimum exists $T(\omega) := \infty$ otherwise. Then $T \colon \Omega \to [0,\infty]$ is measurable. We have \begin{align*} p &:= \P(U_k \le f(V_k)/60)\\ &= \int_{[0,1]^2} \chi_{\{x \le f(y)/60\}} \,d(x,y)\\ &= \int_0^1\int_0^1 \chi_{\{[0,f(y)/60]\}}(x)\, dx\, dy\\ &= \int_0^1 \frac 1{60} f(y)\, dy\\ &= \int_0^1 y^3(1-y)^2\, dy\\ &= \frac 1{60} \left[ 10y^6 - 24y^5 + 15y^4\right]_0^1\\ &= \frac 1{60}. \end{align*} We have therefore \begin{align*} \P(T= k) &= \P\bigl(U_1 > f(V_1)/60, \ldots, U_{k-1} > f(V_{k-1})/60), U_k \le f(V_k)/60\bigr)\\ &= \P(U_1 > f(V_1)/60)\cdots \P(U_{k-1} > f(V_{k-1})/60)\P(U_k \le f(V_k)/60)\\ &= (1-p)^{k-1}p \end{align*} and hence $\P(T < \infty) = 1$. By definition, we have $X = U_T$ and hence, for $x \in (0,1)$: \begin{align*} \P(X \le x) &= \P(U_T \le x)\\ &= \sum_{k=1}^\infty \P(U_T \le x, T = k)\\ &= \sum_{k=1}^\infty \P(U_k \le x, T=k)\\ &= \sum_{k=1}^\infty \P(U_1 > f(V_1)/60, \ldots, U_{k-1} > f(V_{k-1})/60, U_k \le \min\{f(V_{k-1})/60, x\})\\ &= \sum_{k=1}^\infty (1-p)^{k-1} \int_0^x f(y)/60 \, dy\\ &= \frac 1{60p}\int_0^x f(y)\, dy\\ &= \int_0^1 f(y)\, dy. \end{align*} That is, $X$ has density $f$.