Let $F(a,b,c,d) = a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac25$.
With help of a CAS, one can verify
$$\begin{align}
F\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)
&= p^2 - pq + q^2 - qr + r^2 -rs + s^2\\
&= \frac12\left(p^2 + (p-q)^2 + (q-r)^2 + (r-s)^2 + s^2\right)\end{align}$$
If one set $(a,b,c,d)$ to $\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)$, one find
$$\begin{align}
F(a,b,c,d) = 0
&\iff p = p-q = q-r = r-s = s = 0\\
&\iff p = q = r = s = 0
\end{align}
$$
This implies the equation at hand has a unique solution:
$$(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$$
Update
About the question how I come up with this. I first write $F(a,b,c,d)$ as
$$\begin{align}
F(a,b,c,d) &= a^2 + b^2 + c^2 + d^2 - ab - bc - cd - da + d(a-1) + \frac25\\
&= \frac12((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2) + d(a-1) + \frac25
\end{align}\tag{*1}
$$
To simplify the term $d(a-1)$, I introduce $\lambda, \mu$ such that
$$\begin{cases}
d &= \frac12 + \lambda + \mu\\
a &= \frac12 + \lambda - \mu
\end{cases}
\quad\implies\quad d(a-1) = \lambda^2 - \left(\frac12+\mu\right)^2
$$
Now $d-a = 2\mu$ and $(a-b)^2 + (b-c)^2 + (c-d)^2
\ge 3\left(\frac{d-a}{3}\right)^2
= \frac43 \mu^2$.
If one substitute this back into $(*1)$, one find
$$F(a,b,c,d) \ge \frac83\mu^2 + \lambda^2 - (\frac12 + \mu)^2 + \frac25
= \lambda^2 + \frac53\left(\mu - \frac{3}{10}\right)^2$$
In order for $F(a,b,c,d) = 0$, we need
$$\lambda = 0,\quad\mu = \frac{3}{10}
\quad\text{ and }\quad(a-b)^2 + (b-c)^2 + (c-d)^2 = \frac13(d-a)^2$$
The last condition forces $d-c = c-b = b-a = \frac13(d-a)$ and leads to the solution $(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$.
This is a little bit sloppy to describe, so I look at expansion
of $F(a,b,c,d)$ near the solution and obtain a simpler description of $F$ in terms of $p,q,r,s$.
Best Answer
Hint: Multiply by the common denominator to get
$$xi(x+3y)=(3x+4i)(1+yi)$$
Write as real and imaginary parts.
$$x(x+3y)i=3x+4i+3xyi-4y$$
$$x(x+3y)i=(3x-4y)+(3xy+4)i$$
Thus the real and imaginary parts on both sides are equal. This gives:
$$x(x+3y)=3xy+4$$ $$3x-4y=0$$
I think you can solve this system of equations.