Let $f(x,y) = ax^3 + bx^2y + cxy^2 + dy^3$ for $a,b,c,d \in \mathbf Q$. If $a$ or $d$ were $0$ then $f(1,0) = a$ would be $0$ or $f(0,1) = d$ would be $0$. Therefore the assumption that $f$ has no zeros in $\mathbf P^1(\mathbf Q)$ implies $a$ and $d$ are not $0$.
Now consider
$$
F(x) = f(x,1) = ax^3 + bx^2 + cx + d,
$$
which is cubic in $\mathbf Q[x]$ with no rational root: if $F(r) = 0$ for some $r \in \mathbf Q$ then $f(r,1) = 0$, but $f$ has no rational zero. A cubic in $\mathbf Q[x]$ with no root in $\mathbf Q$ is irreducible over $\mathbf Q$ (an accident of small-degree polynomials!), so $F(x)$ is irreducible over $\mathbf Q$. The following theorem explains why $F(x)$ has no root in some $\mathbf Q_p$.
Theorem. If $F(x) \in \mathbf Q[x]$ is irreducible over $\mathbf Q$ and has a root in $\mathbf Q_p$ for all $p$ then $F(x)$ is linear.
The same result and proof hold with $\mathbf Q$ replaced by an arbitrary number field (and the $\mathbf Q_p$'s replaced by all the nonarchimedean completions of that number field), but I stick to coefficients in $\mathbf Q$ to keep the notation simple.
Proof. Let $K = \mathbf Q(\alpha)$ where $F(\alpha) = 0$ and let $L$ be the Galois closure of $K$ over $\mathbf Q$.
The discriminant ${\rm disc}(F)$ is nonzero since $F(x)$ is irreducible over $\mathbf Q$.
Let $S$ be the primes $p$ that appear in the factorization of ${\rm disc}(F)$ or that ramify in $L$ or that appear in a coefficient of $F(x)$.
That is a finite set of primes. For $p \not\in S$, $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable since ${\rm disc}(F) \not\equiv 0 \bmod p$.
For $p \not\in S$, $F(x)$ has a root $r$ in $\mathbf Q_p$ by hypothesis, and the root is in $\mathbf Z_p$ since $F(x)$ has $p$-integral coefficients and its leading coefficient (in fact each coefficient) is a $p$-adic unit.
Therefore we can reduce the equation $F(r) = 0$ to $\overline{F}(\overline{r}) = 0$ in $\mathbf Z_p/(p) = \mathbf F_p$. We have $K \cong \mathbf Q(r)$, so some prime ideal factor of $p$ in $K$ has residue field degree $1$: the embedding of $K$ into $\mathbf Q_p$ by mapping $\alpha$ to $r$ puts a non-archimedean absolute value $|\cdot|_{\mathfrak p}$ on $K$ with residue field $\mathbf F_p$, so $f(\mathfrak p|p) = 1$.
Now we bring in Frobenius elements. Set $G = {\rm Gal}(L/\mathbf Q)$ and $H = {\rm Gal}(L/K)$. Since $p$ is unramified in $L$ (by $p \not\in S$), we can talk about a Frobenius element at a prime $\mathfrak P$ over $p$ in $L$. Since $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable, the permutation action of ${\rm Frob}(\mathfrak P|p)$ on the roots of $F(x)$ has a cycle structure that matches the degrees of the irreducible factors of $F(x) \bmod p$. (The actual choice of $\mathfrak P$ over $p$ doesn't matter, since all Frobenius elements at primes over $p$ are conjugate to each other in $G$ and thus permute the roots of $F(x)$ with the same cycle structure.)
We know $F(x) \bmod p$ has a root in $\mathbf F_p$, so it has a linear irreducible factor mod $p$. Thus ${\rm Frob}(\mathfrak P|p)$ has a fixed point in its permutation action on the roots of $F(x)$: this Frobenius element fixes a root of $F(x)$ in $L$, so this Frobenius element fixes one of the subfields of $L$ conjugate to $K = \mathbf Q(\alpha)$. The subgroup of $G$ corresponding to that subfield is conjugate to $H$. Hence ${\rm Frob}(\mathfrak P|p) \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$.
Now we bring in the Chebotarev density theorem: every element of $G$ is a Frobenius element at infinitely many primes, in fact at a positive density of primes. So each $g \in G$ is a Frobenius element at a prime ideal lying over a prime outside of $S$. The above argument proves $g \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$. Thus
$$
G = \bigcup_{\sigma \in G} \sigma H\sigma^{-1}.
$$
A finite group is never the union of conjugates of a proper subgroup, so $H = G$ and thus $K = \mathbf Q$. Therefore $\mathbf Q(\alpha) = \mathbf Q$, so $F(x)$ has a root in $\mathbf Q$. Since $F(x)$ is irreducible over $\mathbf Q$ with a rational root, $F(x)$ is linear. QED
In this proof, we did not need $F(x)$ to have a root in $\mathbf Q_p$ at every prime $p$. It is okay to say $F(x)$ has a root in $\mathbf Q_p$ outside a set $S$ of primes of density $0$, since Chebotarev implies that each $g \in G$ is a Frobenius at a prime outside $S$. For example, if $F(x)$ has a root in $\mathbf Q_p$ for all but perhaps finitely many $p$ then we can still deduce that $F(x)$ has a root in $\mathbf Q$.
Therefore an irreducible $F(x) \in \mathbf Q[x]$ of degree greater than $1$ has no root in $\mathbf Q_p$ for infinitely many $p$. When $F(x)$ is cubic, not having a root in a field is equivalent to being irreducible over that field. Therefore a cubic irreducible in $\mathbf Q[x]$ has no root in $\mathbf Q_p$ for infinitely many $p$. Returning to the binary cubic form, $f(x,1)$ has no root in $\mathbf Q_p$ of infinitely many $p$, so for infinitely many $p$ we can't solve $f(x,y) = 0$ with $[x,y] \in \mathbf P^1(\mathbf Q_p)$ with $y \not= 0$. Of course we can't solve it with $y = 0$ either, since that would be the point $[x,0] = [1,0]$ and $f(1,0) = a \not= 0$.
I don't think that the Hasse principle is useful here. First, Hasse-Minkowski fails in general for cubic polynomials, e.g., for $3x^3+4y^3+5z^3$. For $x^3+y^3+z^3=n$ I don't see how to use it.
Secondly, Ramanujan has used generating functions to obtain parametrised solutions for $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$, and this direction appears to be more interesting, see for example the article by Harper.
In $1992$, Roger Heath-Brown conjectured that every
$n$ unequal to $4$ or $5$ modulo $9$ has infinitely many representations as sums of three cubes.Conversely it easy to see that the other $n$ cannot be a sum of three cubes. There are many computational aspects, too, which made the problem popular for $n=33$ and $n=42$.
Best Answer
By a theorem of Hilbert, Hasse principle holds for the affine cubic form $x\mapsto N(x)-a$ ( a rational number is globally a norm of a cyclic extension if and only if it is locally)
Note this is true if you replace $3$ by any integer, provided you restrict yourself to cyclic extensions. You can also replace $\mathbb{Q}$ ny a global field.
You can find a nice introduction to the definition here: http://homepages.warwick.ac.uk/staff/S.Siksek/arith/notes/brauermanin.pdf
For example, Bayer, Lee and Parimala computed the Brauer Manin obstruction for multinorm equations, and I'm pretty sure that we can use their work to construct counterexamples of arbitrary large degree. The full paper is available on Parimala's web page.