= Solution
For a thin wire carrying current $I$, the given <magnetic vector potential> reduces to $\mathbf A=(\mu_0I/4\pi)\oint_Cd\mathbf r'/|\mathbf r-\mathbf r'|$. Taking its <curl> and using $\nabla_{\mathbf r}(1/|\mathbf r-\mathbf r'|)=-(\mathbf r-\mathbf r')/|\mathbf r-\mathbf r'|^3$ gives the <Biot-Savart law>
$$
\boxed{\mathbf B(\mathbf r)=\frac{\mu_0I}{4\pi}\oint_C\frac{d\mathbf r'\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}.}
$$
This is exactly the sign convention in the printed form using $\mathbf r'-\mathbf r$.
Let $\mathbf v=(\mathbf r-\mathbf r')/|\mathbf r-\mathbf r'|^3$. Away from the wire, $\nabla\cdot\mathbf v=0$. The supplied <divergence and curl of a cross product> therefore gives $\nabla\times(d\mathbf r'\times\mathbf v)=-(d\mathbf r'\cdot\nabla_{\mathbf r})\mathbf v$. But $\nabla_{\mathbf r}\mathbf v=-\nabla_{\mathbf r'}\mathbf v$, so the integrand is the total differential of $\mathbf v$ along the source loop. Its closed-loop integral is zero. Thus \b[$\nabla\times\mathbf B=0$ at every point outside $C$], as demanded by the current-free magnetostatic <Ampère's law>.
For $r'=f(\theta)>0$, $z'=0$, and counterclockwise current, $d\mathbf r'=(f'\mathbf e_r+f\mathbf e_\theta)d\theta$. At the origin, $\mathbf v=-\mathbf e_r/f^2$, so
$$
\boxed{\mathbf B(0)=\widehat{\mathbf z}\,\frac{\mu_0I}{4\pi}\int_0^{2\pi}\frac{d\theta}{f(\theta)}.}
$$
For the <ellipse> expressed relative to a focus, $1/f=(1-e\cos\theta)/\ell$. The cosine integrates to zero, giving
$$
\boxed{\mathbf B_{\mathrm{focus}}=\widehat{\mathbf z}\,\frac{\mu_0I}{2\ell}.}
$$
Here $\ell$ is the <semilatus rectum>. Reversing the current reverses the field direction.
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