= Solution
Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim0$. The <Riemann-Roch theorem> therefore gives
$$
\ell(nQ)=n\qquad(n\geq1).
$$
Choose
$$
x\in L(2Q)\setminus L(Q),
\qquad
y\in L(3Q)\setminus L(2Q).
$$
The <rational function>[rational functions] $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.
The seven functions
$$
1,x,y,x^2,xy,x^3,y^2
$$
belong to the six-dimensional <Riemann-Roch space> $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the <Weierstrass equation of an elliptic curve>[Weierstrass form]
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.
$$
The divisor $3Q$ is <very ample divisor>[very ample], so its complete linear system embeds $E$ in $\mathbb P^2$ and the displayed relation cuts out its image; this is the <Weierstrass construction from Riemann-Roch in genus one>.
Because $\operatorname{char}k\ne2$, completing the square removes the terms linear in $y$, and because $\operatorname{char}k\ne3$, translating $x$ removes the quadratic term. Thus the equation becomes
$$
y^2=x^3+Ax+B.
$$
Since $k$ is <algebraically closed field>[algebraically closed], the cubic factors as
$$
x^3+Ax+B=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3).
$$
Smoothness of $E$ says that the <elliptic-curve discriminant> is nonzero, equivalently the three roots are distinct. Hence
$$
\boxed{E\cong\{y^2=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)\}\subset\mathbb P^2.}
$$
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