Python 2, 40 39 bytes.

f=lambda x:x<4or.0+f(x-3)+f(x-2)/f(x-1)

Gives True instead of 1, if this isn't allowed we can have this for 42 bytes:

f=lambda x:.0+(x<4or f(x-3)+f(x-2)/f(x-1))

The way it works is pretty straightforward, the only trick is using .0+ to cast the result to a float.

Ruby, 34 bytes

Since Ruby uses integer division by default, it turns out that it's shorter to use fractions instead. Golfing suggestions welcome.

f=->n{n<4?1r:f[n-3]+f[n-2]/f[n-1]}

C, 46 bytes

float P(n){return n<4?1:P(n-3)+P(n-2)/P(n-1);}

Ideone

Haskel, 32 bytes

(a#b)c=a:(b#c)(a+b/c)
((0#1)1!!)

Usage example: ((0#1)1!!) 7 -> 2.642857142857143. I start the sequence with 0, 1, 1 to fix !!'s 0-based indexing.

Edit: @xnor found a way to switch from 0-based to 1-based index, without changing the byte count.

MATL, 15 bytes

llli3-:"3$t/+]&

Try it online!

Explanation

lll       % Push 1, 1, 1
i         % Take input n
3-:       % Pop n and push range [1 2 ... n-3] (empty if n<4)
"         % For each
  3$t     %    Duplicate the top three numbers in the stack
  /       %    Pop the top two numbers and push their division
  +       %    Pop the top two numbers and push their addition
]         % End
&         % Specify that the next function, which is implicit display, will take
          % only one input. So the top of the stack is displayed

Perl 6,  25  23 bytes

{(0,1,1,1,*+*/*...*)[$_]}

{(0,1,1,*+*/*...*)[$_]}

Explanation:

# bare block lambda with implicit parameter ｢$_｣
{
  (
    # initial set-up
    # the ｢0｣ is for P(0) which isn't defined
    0, 1, 1, 1,

    # Whatever lambda implementing the algorithm
    * + * / *
    # { $^a + $^b / $^c }

    # keep using the lambda to generate new values until
    ...

    # Whatever (Forever)
    *

   # get the value indexed by the argument
  )[ $_ ]
}

This returns a Rat (Rational) for inputs starting with 3 up until the result would start having a denominator bigger than can fit in a 64 bit integer, at which point it starts returning Nums (floating point).
The last Rat it will return is P(11) == 8832072277617 / 2586200337022

If you want it to return Rational numbers rather than floats you can swap it for the following which will return a FatRat instead.

{(0.FatRat,1,1,*+*/*...*)[$_]}

Test:

#! /usr/bin/env perl6
use v6.c;
use Test;

my &piggyback = {(0,1,1,*+*/*...*)[$_]}
# */ # stupid highlighter no Perl will ever have C/C++ comments

my @test = (
  1, 1, 1, 2,
  3/2, 7/3, 37/14,
  529 / 222,
  38242 / 11109,
  66065507 / 19809356,
  8832072277617 / 2586200337022,
);

plan +@test;

for 1..* Z @test -> ($input,$expected) {
  cmp-ok piggyback($input), &[==], $expected, $expected.perl;
}

Cheddar, 31 bytes

n P->n<4?1:P(n-3)+P(n-2)/P(n-1)

The ungolfed version is so clear imo you don't need explanation:

n P->
  n < 4 ? 1 : P(n-3) + P(n-2) / P(n-1)

basically after the function arguments you can specify the variable to use which will be set to the function itself. Why? because this function will be tail-call-optimized, or at least should be.

Pyth, 20 bytes

L?<b4h0+y-b3cy-b2ytb

Try it online!

Javascript (ES6), 31 bytes

P=n=>n<4?1:P(n-3)+P(n-2)/P(n-1)

A simple function.

P=n=>n<4?1:P(n-3)+P(n-2)/P(n-1)

var out = '';

for (var i=1;i <= 20;i++) {
out +='<strong>'+i+':</strong> '+P(i)+'<br/>';
}

document.getElementById('text').innerHTML = out;

div {
font-family: Arial
}

<div id="text"></div>

Jelly, 15 bytes

ạ2,1,3߀÷2/SµḊ¡

Try it online! or verify all test cases.

How it works

ạ2,1,3߀÷2/SµḊ¡  Main link. Argument: n (integer)

             Ḋ   Dequeue; yield [2, ..., n].
            µ ¡  If the range is non-empty (i.e., if n > 1), execute the chain to
                 the left. If n is 0 or 1, return n.
                 Note that P(3) = P(0) + P(2)/P(1) if we define P(0) := 0.
ạ2,1,3           Take the absolute difference of n and 2, 1, and 3.
                 This gives [0, 1, 1] if n = 2, and P(0) + P(1)/P(1) = 0 + 1/1 = 1.
      ߀         Recursively apply the main each to each difference.
        ÷2/      Perform pairwise division.
                 This maps [P(n-2), P(n-1), P(n-3)] to [P(n-2)/P(n-1), P(n-3)].
           S     Sum, yielding P(n-2)/P(n-1) + P(n-3).

05AB1E, 18 17 bytes

3Ld                # push list [1,1,1]
   ¹ÍG         }   # input-3 times do
      D3£          # duplicate list and take first 3 elements of the copy
         R`        # reverse and flatten
           /+      # divide then add
             ¸ì    # wrap in list and prepend to full list
                ¬  # get first element and implicitly print

Try it online!

Saved 1 byte thanks to Luis Mendo

Mathematica, 36 bytes

P@n_:=If[n<4,1,P[n-3]+P[n-2]/P[n-1]]

Here are the first few terms:

P /@ Range[10]
{1, 1, 1, 2, 3/2, 7/3, 37/14, 529/222, 38242/11109, 66065507/19809356}

Dyalog APL, 25 bytes

⊃{1↓⍵,⍎⍕' +÷',¨⍵}⍣⎕⊢0 1 1