= Solution
Let $P(x)=\prod_{i=1}^r(x-\lambda_i)$. Since both $x$ and $y$ are invertible on $X$, the formula
$$
F(x,y)=x+\frac1{xy}
$$
defines a regular map $F:X\to\mathbb A^1$.
Fix $t\in k$. To solve $F(x,y)=t$, set
$$
y=\frac1{x(t-x)}.
$$
The curve equation becomes
$$
x^2(t-x)^2P(x)-1=0.
$$
This is a nonconstant polynomial, so it has a root because $k$ is algebraically closed. At a root, neither $x$ nor $t-x$ can vanish, and the equation also forces $P(x)\ne0$. The displayed value of $y$ is therefore defined and nonzero, satisfies $y^2=P(x)$, and obeys $F(x,y)=t$. Thus $F$ is surjective.
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