= Solution
Write an integral <binary quadratic form> as $[a,b,c]=ax^2+bxy+cy^2$, with <discriminant of a binary quadratic form> $D=b^2-4ac$. It is positive definite when $a>0$ and $D<0$. Two forms are <proper equivalence of binary quadratic forms>[equivalent] if one is obtained from the other by an integral linear change of variables whose matrix lies in $\operatorname{SL}_2(\mathbb Z)$. A positive definite form is <reduced positive definite binary quadratic form>[reduced] when
$$
|b|\leq a\leq c,
$$
with $b\geq0$ when $|b|=a$ or $a=c$.
To reduce a form, apply a unimodular shear $x\mapsto x+ny$ with $n$ chosen so that the new middle coefficient has $|b'|\leq a$. If the new final coefficient $c'$ is smaller than $a$, exchange the variables, using a determinant-one rotation, so that the positive leading coefficient decreases from $a$ to $c'$. Repeating must stop because a positive integer cannot decrease indefinitely. At termination $|b|\leq a\leq c$, and elementary sign changes impose the boundary convention. This is the <reduction algorithm for a positive definite binary quadratic form>, and proves that every positive definite form is equivalent to a reduced one.
The displayed forms have discriminant $1-36=-35$, so the intended discriminant is $-35$. For any reduced form of this discriminant, <enumeration of reduced binary quadratic forms> gives
$$
1\leq a\leq\sqrt{\frac{35}{3}}<4,\qquad |b|\leq a,
\qquad c=\frac{b^2+35}{4a}\in\mathbb Z.
$$
The boundary convention and direct checking of $a=1,2,3$ leave only
$$
\boxed{f_1=[1,1,9]=x^2+xy+9y^2,\qquad
f_2=[3,1,3]=3x^2+xy+3y^2.}
$$
Suppose first that a prime $p>7$ is represented by $f_1$. The identity
$$
4f_1(x,y)=(2x+y)^2+35y^2
$$
shows modulo both $5$ and $7$ that $4p$ is a nonzero square. Hence
$$
\left(\frac p5\right)=\left(\frac p7\right)=1.
$$
Similarly,
$$
12f_2(x,y)=(6x+y)^2+35y^2.
$$
Modulo $5$, this says $2p$ is a square, while modulo $7$ it says $5p$ is a square. Since $(2/5)=(5/7)=-1$, the <multiplicativity of the Legendre symbol> gives
$$
\left(\frac p5\right)=\left(\frac p7\right)=-1.
$$
Conversely, suppose the two displayed <Legendre symbols> have the same sign. By <quadratic reciprocity>,
$$
\left(\frac{-35}{p}\right)
=\left(\frac p5\right)\left(\frac p7\right)=1.
$$
Choose an odd integer $B$ satisfying $B^2\equiv-35\pmod p$. Since $B$ is odd, $B^2+35$ is also divisible by $4$, so
$$
F=\left[p,B,\frac{B^2+35}{4p}\right]
$$
is a positive definite integral form of discriminant $-35$ that represents $p$. Reducing $F$ yields either $f_1$ or $f_2$, and equivalence preserves represented integers. The necessary symbol calculation above determines which one: the positive pair selects $f_1$, and the negative pair selects $f_2$. Therefore
$$
\boxed{
p\text{ is represented by }f_i
\iff
\left(\frac p5\right)=\left(\frac p7\right)
=
\begin{cases}
+1,&i=1,\\
-1,&i=2.
\end{cases}}
$$
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