We want to prove that $\forall\epsilon>0,\forall\delta>0$ if $\rVert(x,y)\rVert<\delta$, then $\bigg|\frac{\cos (xy) - 1}{x}\bigg|<\epsilon$. Let's stipulate $\delta=:\epsilon/3$. Focusing on the numerator of the expression we want to bound, we get
$$|\cos(xy)-\cos^2(xy)-\sin^2(xy)|\leq |\cos(xy)-\cos^2(xy)| + |\sin(xy)|^2 \\ \leq|\cos(xy)||2|\bigg|\frac{1-\cos xy}{2}\bigg| + |\sin(xy)| \leq |\cos(xy)||2||\sin^2(xy/2)| + |\sin(xy)| \leq 2|xy|$$
Hence
$$\bigg| \frac{\cos (xy) - 1}{x} \bigg| \leq \bigg|\frac{2xy}{x}\bigg| = 2|y|< 2\delta < \epsilon $$
Notice that
$$ \frac{1 - \cos(xy)}{x} = \int_0^y \sin(xt)dt = y \int_0^1 \sin(xyt')dt'$$
so we have
$$\left|\frac{\cos(xy)-1}{x}\right| = |y|\left|\int_0^1 \sin(xyt)dt\right| \leq |y|\int_0^1 |\sin(xyt)|dt$$
$$\leq |y|\int_0^1 |xyt|dt \leq |x|y^2$$
From here using $\delta^3 = \epsilon$ yields the desired inequality.