Do you hate square roots?

$\Rightarrow x\sqrt{y}=30-y\sqrt{x}$

$\color{#3D99F6}{x\sqrt{y}+y\sqrt{x}}=30$

$\Rightarrow x\sqrt{x}=35-y\sqrt{y}$

$x\sqrt{x}+y\sqrt{y}=30+5$

$x\sqrt{x}+y\sqrt{y}=(\color{#3D99F6}{x\sqrt{y}+y\sqrt{x}})+5$

$x\sqrt{x}-y\sqrt{x}+y\sqrt{y}-x\sqrt{x}=5$

$\sqrt{x}(x-y)-\sqrt{y}(x-y)=5$

$(\sqrt{x}-\sqrt{y})(x-y)=1×5$

Now by comparing,
We get $x=9$ and $y=4$ .

Putting these values in $\sqrt{x^2+y^2+3}$ .

$\sqrt{(9)^2+(4)^2+3}=\sqrt{100}=\boxed{10}$

$\begin{cases} x \sqrt{y} = 30 - y \sqrt{x} & ...(1) \\ x \sqrt{x} = 35 - y \sqrt{y} & ...(2) \end{cases}$

$\begin{aligned} (1): \quad \quad \quad \quad x\sqrt{y} & = 30 - y\sqrt{x} \\ x\sqrt{y} + y\sqrt{x} & = 30 \quad \quad \quad \quad \small \color{#3D99F6}{\text{Let } a=\sqrt{x} \text{ and } b=\sqrt{y}} \\ a^2b + ab^2 & = 30 \\ \Rightarrow \color{#D61F06}{ab} \color{#3D99F6}{(a+b)} & = 30 \quad \quad \quad \quad \small \text{Let } \color{#3D99F6}{\alpha = a+b} \text{ and } \color{#D61F06}{\beta = ab} \\ \color{#3D99F6}{\alpha} \color{#D61F06}{\beta} & = \color{#3D99F6}{30} \end{aligned}$

$\begin{aligned} (2): \quad \quad \quad \quad \quad \quad \quad x\sqrt{x} & = 35 - y\sqrt{y} \\ x\sqrt{x} + y\sqrt{y} & = 35 \\ a^3 + b^3 & = 35 \\ (a+b)(a^2+b^2-ab) & = 35 \\ (a+b)[(a+b)^2 - 3ab] & = 35 \\ \alpha^3 - 3 \color{#3D99F6}{\alpha} \color{#D61F06}{\beta} & = 35 \\ \alpha^3 - 3 (\color{#3D99F6}{30}) & = 35 \\ \Rightarrow \alpha^3 & = 125 \\ \alpha & = 5 \\ \Rightarrow \beta & = 6 \end{aligned}$

Now we have:

$\begin{aligned} \sqrt{x^2+y^2+3} & = \sqrt{a^4+b^4+3} \\ & = \sqrt{(a+b)(a^3+b^3)-ab(a^2+b^2)+3} \\ & = \sqrt{35\alpha - \beta(\alpha^2 -2\beta) +3} \\ & = \sqrt{35(5) - 6(5^2 -2(6)) + 3} \\ & = \sqrt{100} \\ & = \boxed{10} \end{aligned}$

Rewrite the 2 equations:

$x\sqrt { y } +y\sqrt { x } =30$

$x\sqrt { x } +y\sqrt { y } =35$

Square both equations:

${ x }^{ 2 }y+2xy\sqrt { xy } +x{ y }^{ 2 }={ 30 }^{ 2 }$

${ x }^{ 3 }+2xy\sqrt { xy } +{ y }^{ 3 }={ 35 }^{ 2 }$

Subtract: ${ x }^{ 3 }-{ x }^{ 2 }y+{ y }^{ 3 }-{ y }^{ 2 }x={ 35 }^{ 2 }-{ 30 }^{ 2 }=325$

Grouping: $(x-y)({ x }^{ 2 }-{ y }^{ 2 })=(x+y){ (x-y })^{ 2 }=325$

The only square factors of 325 are 1 and 25, but 1 will not work. So this diophantine equation can only have 2 solutions that can be yielded from:

$x+y=13$

$x-y=\pm 5$

Adding and subtracting the equations ,we get

$(x - y) (\sqrt x - \sqrt y) = 5$ and $(x + y) (\sqrt x + \sqrt y) = 65$

Multiplying them, $(x^2 - y^2)(x - y) = 65\times 5$

Let, $(x^2 - y^2) = 65$ and $(x - y) = 5 \Rightarrow x = y + 5$

$(y + 5)^2 - y^2 = 65\Rightarrow y = 4$ Therefore, $x = 4 + 5 = 9$

$\sqrt{x^2 + y^2 + 3} = \sqrt{9^2 + 4^2 + 3} = \sqrt{100}= \boxed{10}$

$For~ease~of~calculations,~let~a=\sqrt x,~~~and~~~b=\sqrt y.\\ \therefore~The ~two~equations~are:-\\ (A)~~~a^3+b^3=35,~~~~~~~~~~(B)~~~ab(a+b)=30.\\ And~(C)~~~\sqrt{x^2+y^2+3}=\sqrt{a^4+b^4+3}.\\ From~(A)~~~a^3+b^3=27+8=3^3+2^3=35.~~~\therefore~a=3~~and~~b=2.\\ Substituting~in~(B),~~LHS=3*2(3+2)=30=RHS.\\ So~a~IS~~ EQUAL~~TO~3~~and~~b~IS~~EQUAL~~TO~2.\\ \implies~\sqrt{a^4+b^4+3}=\sqrt{81+16+3}=10.\\ From~~(C)~~\sqrt{x^2+y^2+3}=\Large~~~\color{#D61F06}{10}.$

Its all quite simple if you can visualise what exactly it needs. Though its a hit and trial but its works quite often. Lets take x=9 and Y = 4. Its statifies both the equations and hence, The answer is 10.